Apologies for the re-upload. In the previous version I only solved for positive integers. It is straightforward to then find the other solution. But I uploaded this corrected video presenting both because not everyone reads the comments. Also these videos are getting shared in a lot of places so I'm doing my best to present accurate mathematical results. I thank Aniket Gupta and Ryan Wilson for pointing out the oversight!
So where did I go wrong in the original video? I often get emails where people ask me to look for mistakes in their work, if they did not find the correct solution in one of my videos. One important skill in mathematics is reviewing your own work. It's not always fun, but it's really important to learn where you might make errors. So where did I go wrong? In my self-review, there are three places I could have realized my mistake. Place #1: I said the difference of factors is (2^n + x) - (2^n - x) = 2x. Then I solved 2x = 118 to get x = 59. The thing is the difference of factors can also be (2^n - x) - (2^n + x) = -2x. Setting that equal to 118 would have given x = -59. Place #2: When I solved y = 12, I could have then solved 615 + x^2 = 2^12 where I could have seen x = 59 or -59. Place #3: In checking my work, I should have seen there was an x^2 term, and it's a common mistake to overlook a negative solution. This is the kind of self-criticism I do for my math homework. I have noticed people don't like to admit if they are wrong. But I have found self-reviews and honesty are big strengths to improving math skills. So I thank all the legitimate criticism (if accompanied with legitimate honesty) for these videos--you're helping us all get better at math.
@BlazePlayz YT of course you can just blindly type in numbers, that´s easy, but what if the answer was smth like y=22 i think you would have some troubles then
Rishi Jai is there a certain trick to spotting a high number with a square root? Cus I did the blindly squaring numbers on my calculator around the sums + 615 being a power of 2 thing...
It is easier to solve this problem with little logical way than to follow Presh`s complicated method. Look at the problem once. 2^y is always an even number irrespective of whether y is even or odd. Now we can write the equation as 615 =2^y - x^2. Here 615 is an odd number. Therefore, x^2 will have to be an odd number. Because, the difference between two even numbers is always even and the difference between an even number and an odd number is always odd. Since x^2 is an odd number, x must be an odd number too as the square of an odd number is always an odd number and the vice-versa is also true. Let x = 2n+1 for any value of n, odd or even. Now we have 2^y - (2n+1)^2 = 615. This can be written as (2^(y/2))^2 - (2n+1)^2 = 615. This now becomes the difference between two squares form. That means the left hand side will be the product of [(2^(y/2)) + (2n+1)][2^(y/2) - (2n+1)]. Putting, 2^(y/2) = a, and 2n+1 = b for simplification, we have (a+b)(a-b) = 615. Now 615 has to be the product of two numbers. The number 615 = 1 x 3 x 5 x 41. Therefore, we have four alternative pair of numbers whose product will be 615. They are (i) a+b = 615 and a - b = 1 or (ii) a + b = 205 and a - b = 3 or (iii) a + b =123 and a - b = 5 or (iv) a + b = 41 and a - b = 15. Solving (i) we have a = 308, b = 307 (ii) a =104, b = 101 (iii) a = 64, b = 59 and (iv) a = 28, b = 13. Here note that 'a' = 2^(y/2) that is a number power of 2. If we check the values of 'a' in all thee alternatives, only (iii) which is 64,satisfies our condition and the others are not.Therefore 2^(y/2) = 64 = 2^6. That is y/2 =6, That is y = 12 and b = 2n+1 = x =59.
@@rcnayak_58 you don't even have to set b = 2n+1 because you don't use it again. Also, xmarteo's comment raises a valid concern, because then you would have to check for more divisors (divisors on Z[sqrt(2)] ). Or you could show that y is even, resulting in the same proof as MindYourDecisions. Ultimately, your proof is not shorter than MindYourDecisions'.
@@rcnayak_58 Try your logic with 23 + x^2 = 2^y. You'll get that (a+b)(a-b) = 23, which only has one integer solution: (a, b) = (12, 11), and since 12 is not a power of 2, you would conclude that there's no solution with integer x and y. But (x, y) = (3, 5) is a solution, corresponding to (a, b) = (4sqrt(2), 3).
@@xmarteo 2^y is always even because it's a multiple of 2 and an even number is defined as multiple to, so by definition 2^y is even. The problem though, is that this solution is almost as complex as Presh's, because it practically requires the same methodology of thinking. While I agree it might be efficient, you've to keep in mind that he was send this solution as well.
You can solve it beautifully in Excel by solving (with parameters) an equation as 615+x^2 - 2^y = 0 Alternatively you can calculate y=log(615+x^2)/log(2) and then an integer value of y should be found. All to be solved in Excel :)
I'd give you a cookie, but you only really deserve it if you use SQL. (By the way, good luck brute-forcing the proof that there are no other solutions.)
After spotting that n > 9, I just used trial and error with n = 10, 11... to check which values would give a perfect square for 2^n - 615. Eventhough, trial and error isnt the 'best' way to solve problems, I think this would be a quicker approach :) THOUGH I ABSOLUTELY LOVED THE WAY YOU DID IT, IT IS ALWAYS FUN TO KNOW HOW THE OTHER PERSON SOLVED IT. I LOVE MULTIPLE APPROACHES
Hey presh. Just wanted to thank you... we had a really difficult math test that was more about thinking than calculating. It was really hard but everytime i started a new question, I thought about what Presh will do and how he will solve that, and that helped solving most of the problems. Thanks for improving and sharpening the thinking for me, and for other hundreds of thousands people
@@kwcy92 Of course! It was a guessing game with math background 😂😂😂 We can rewrite the problem as 615=2^y-x^2 I know x will be an odd number (because 615 is odd and 2^y is always even). Because we know powers of two, we want a “bigger” number for y because difference needs to be 615 From there on I played a bit with numbers to see what happens when I use bigger x, smaller y... I started with x=y=9 and moved up by 5-10 for x and one for y and got there relatively fast. I know it’s not the best way to solve, but it worked 😂😂😂
615 is divisible by 3, hence 2^y-x^2 is divisible by 3. Easily checking out we could see that 2-x^2 is not divisible by 3, then 2^y-0 or 2^y-1 is divisible by 3. Since 2^y-0 isn't, we can conclude that 2^y-1 is, then y is even
I also started as Nayak's solution: x has to be odd, then x=2n+1. Then, we have that 4n(n+1)=2^y-616, or n(n+1)=2^(y-2)-154. Now I went with y-2>=8, and found y-2=10 and n=29, which give x=59 and y=12. Cheers, Marco.
I always depend on Wolfram Alpha for solving Equations in Physics. But this new information blows my mind. Now I have to go back and do all those tedious exercises from Goldstein.
For mathematics, only use wolfram for getting hints toward a solution. Often wolfram does not give you all possible solutions or even managed to find them (this has happened to me a lot in calculus and even combinatorics). Try to understand the problem and the theory behind it and solve it from there. There are lots of mathematics that are not implemented yet in wolfram either.
@mindyourdecisions I took a game theory class in college this past semester (spring 2018) and you would not believe the amount of times you were listed as a source.
Great and imaginative solution! Lots of people commenting on these problems don’t understand mathematics though. They believe one only has to do trial and error to find one solution, but forget all about finding every possible solution while also showing that there can be no other. This is the core thinking that should be taught. Wonder what they actually teach people in school nowadays.
The solution to this problem really blowed my mind. The structure of it is the most imaginative and trickery of what I've encountered so far 😱 Thanks for sharing
In this video, he looks mod 10 to get that y is even. But it is much easier to look mod 4 to get that fact. And also, a lot of people are saying they solved this by guessing and checking. While that does find a solution, it does not find a proof of the only solutions, which this video does.
Neat. I used the mod 10 solution myself, which seemed more obvious to me since I count in base 10 and tend to notice base-10 patterns before considering other bases. (OK, I lie. I did briefly look at mod 2 first, but that only got me as far as "x must be odd" before I abandoned that approach and jumped to base 10. Mod 3 and 4 never occurred to me.)
One more analysis, that 2^y is always even so sum of of 615 and x^2 should also be even, which is possible when x^2 is odd. x^2 is odd when x is odd integer. Hence odd integral solutions of X is only possible.
Take the logarithm base 2 of both sides: Answer: y = log(x^2 + 615)/log(2) + ((2 i) π n)/log(2) for n element Z x = ± 59, y = 12 (that's a Wolfram's answer)
I‘ve been taking some number theory classes in my free time and we‘ve just started with modular arithmetics, and I was so happy when I got the solution before watching the video haha :) I did it mod 3 tho
I solved this in about 60 seconds from the thumbnail. The right hand side has to be 1024 ... 2048 ... 4096 etc. So just keep trying until you find one which has an integer square root. Bingo - square root of 4096 - 615 is 59 (and -59). A familiarity with well known binary numbers helps. Because I solved it before I opened the video the no calculator rule didn’t apply. 🙂
Well how do you say? There is no any proper explanation about proof. It's so abstract and we never realized we need to use inequalities for this kind of problem. You know in HS we always use regular method, but not with this long and tiring explanation only to find y=2n. How come common people realized that? Me as a Master degree student even cannot realize this
nonesuch27 Most people don't grasp the concept of proof. Proof is really only needed in the study of pure mathematics. For everyday applications of mathematics, you don't need proofs, just like you don't need a degree in automotive engineering to be an auto mechanic.
What? What does proof have to do with anything? The problem as worded says nothing about proof. It asks you to find x and y, only allowing integers. Period.
@@qc1okay If it asks to SOLVE the problem then it means you have to find ALL the solutions. If you find only ONE solution, you have to prove its the only solution. Otherwise you didn't solve the problem. Many people just say "easy, i guesed it in 3 tries", while that worked in this case because there's only one solution, it won't always work.
@@dariobarisic3502 No it doesn't. Just that simple. Solve does not mean all. This TH-cam channel isn't a formal math class, and even if it were, its instructions must be clear to its viewers. If "all" were wanted, "all" had to be stated. Incidentally, I went thru the full process Presh used, and it both leaves out steps and overcomplicates others.
Here's how I solved this with paper and pencil. Rewrote expression as x^2 -2^y=-615. X^2 will always be a positive integer value, so 2^y must be some power of 2 greater than 615 where the difference between the two expressions has a (positive) integer square root. 2^10 = 1024 & square root of the difference (409) is not an integer, 2^11=2048 and square root of the difference (1433) is not an integer, 2^12=4096, subtracting 615 yields 3481. Since the last two digits of 3481 are 81, there is a good chance that 59 could be the square root. I tried that and it worked. Thus, x=59, y=12.Yes, I know intuition and trail and error aren't proper 'solutions.' Thanks for this channel!
I solved this in about 3 minutes by simply listing the powers of two, subtracting 615, and looking for an integer square root. The first one even possible had to be more than 615, so I only had to check three powers of two before finding one with an integer square root.
2^y , which means it could only be2,4,8,16,32... Therefore,just to find out which 2^y number can be square rooted after minus 615.There is 615, so we can start with 1024,2048,4096... then find out that the answer is x=59,y=12
I did trial and error method and solved it in 2mins.....2^9=512 & 2^10=1024, 2^y has toh be greater than 615, so consider 2^10 =1024, so 1024-615=409 which is not a Perfect Square, 2^11=2048, so 2048-615=1433 which is also not a Perfect Square, and then 2^12=4096, so 4096-615=3481 which is Square of 59. Thus x=59 and y=12
I just did guess and check and found 59 as a solution for x. Obviously 2^y has to be greater than 615, so I started with 1024-615, which wasn't a perfect square, so I did 2048-615, then 4096-615. I found the square root to be 59. So -59 and 59 are solutions when y=12. I imagine there are infinite solutions. EDIT: The solution in the video is much more elegant, and also shows there is only one solution. Very cool.
I have to admit, when I first saw this problem I completely did not think it would be very complicated. Then I tried it and got stuck but I was still convinced there some basic algebra trick that I'm missing. Then after watching this solution, I realize there was no chance I was solving this without the explanation
I solved it partly using logic and finished it using trial and error, but your method is so much better because it demonstrates the logic required to prove it is the only solution. I look forward to sharing this with my senior students.
Sometimes you come across a map problem and you think “I get it” and you “solve” the problem under a couple of minutes only to realize later the problem was above your “pay grade” and a much more complicated problem that you anticipated. This has happened to me back to back today ;)
@@TechToppers not realy that hard. Every square is 0 or 1 mod 3, every even power of 2 is 1 mod 3, but every odd power of 2 is 2 mod 3. So the y is even. You would need to prove the initial statements though (but it's very simple).
@@TechToppers I am not arguing that the solution in the video is very demonstrative. Just saying that going for the reminder of a division is not that hard either.
Who else used a different strategy, but found a much faster way to do it? My strategy was to try different powers of 2 (greater than 9). For example: 615 + x^2 = 2^11 (2048). This implies that x^2 = 1433. That isn't a square value (for a positive even integer), so I tried 615 + x^2 = 2^12 (4096). That meant x^2 was equal to 3481. Since 3481 is a square, I was able to solve for x and y.
This was a very fine problem! It was good that 615 is quite easy to divide into prime factors, otherwise it wouldn't have been so easy without a calculator :)
I solved this directly by the substitution 2 to the yth power = a squared, which gets to the answer more quickly, although the remainder of the approach is similar to what was presented. The above approach leads immediately to (a-x)(a+x) = 615 and by substituting the four ways to factor 615 (41x 15, 123 x 5, 205 x 3, and 615 x 1), the only pair of factors that produce integer solutions is 5 and 123. a - x = 5 and a + x = 123 leads to a = 64 and x = 59, and since a squared = 2 to the yth power, 2 to the yth power must equal 4096, so y = 12.
@Presh, had you constructed the proof the other way around, it would have been a lot closer to how people/ students think. Start with observing x^2 and that if y is even then you can do a^2 - b^2 = 615 which is easy to solve. Then prove that y cannot be odd, thus the only solutions are the ones where y is even (x = +/- 59, y = 12). The way you started it by observing y has to be even and then stating ("This will be a key in our finding the solution") makes you look insightful, but does not help people develop thinking patterns in math problem solving (especially Diophantine equations)
Well... I solved the problem the exact same way and in the exact same order. It's not really about being insightful but about using the fact integers are a strong constraint on the solutions.
@Dan It's the same with last digit analysis. How do you know it is we need in this specific question? You don't know, but i bet factoring leads to results more often than last digit analysis.
If you look at the equation in mod 3 it is clear that y is even which means that you can move x^2 to right side of the equation and do difference of two squares. And then try for positive divisors of 615
I just started at 1024 and subtracted 615 from each power of 2 from there on. 3rd try, 4096. Obviously if it was something like 2^2000, that would take far too long, but for this one...worked in like 20 seconds. 1024 - 615 = 409 (not a square) 2048 - 615 = 1433 (not a square) 4096 - 615 = 3481 (59^2) I think this is why I hate traditional math problems, they make them simple enough that you can brute force an answer instead of doing the work. Got me through all my trig/calc classes in high school :D
Tom Domenge sure, (mod 3) 615 = 0, x^2 = 1 or 0, and 2^y = (-1)^y. So 615 + x^2 = 0 or 1. If y is odd, 2^y=(-1)^y = -1 which is impossible. So y is even
Minha solução: Se "==" representa congruência, temos: 615+x²==x² (mod 3). 2ⁿ=/=0 (mod 3) →x²=/=0 (mod 3) →x²==1 (mod 3). 2ⁿ==1 (mod 3) →n=2a 615=2²ª-x²→ 615=(2ª+x)(2ª-x). Edit: ok, vamos testar todos os pouquíssimos casos: 615=3•5•41, além disso, 2ª+x>2ª-x. As únicas possibilidades são (supondo x≥0): 2ª+x=615 e 2ª-x=1→2•2ª=616. Abs 2ª+x=205, 2ª-x=3→2•2ª=208. Abs 2ª+x=41 e 2ª-x=15→ 2•2ª=56. Abs E 2ª+x=123 e 2ª-x=5 2ª=64, de onde segue que a única solução é a=6→n=y=12 e x=±59
Translation for english: My solution is: If "==" represents congruence then we have: .... Testing all the possible cases (which are just a few) , we have that the only solution is...
To chekc whether y is even or odd we can simply use 3 as x^2 gives remainder 0 or 1 when diveded by 3 and 2^y gives remainder 2 when y is odd and 1 when is even therefore y must be even
For the ones using a computer or a calculator, once you observe that y has to be even, therefore 2^y is a square. Notice that once you're at a perfect square n², you can only get to another square by substracting 615 only as long as n is small? How small you ask, (n+1) ² - n² > 615, this gives n>307. Therefore once 2^y exceeds 307² , there's no point checking. Therefore you only need to check for the values of y= 10,12,14,16
This is so cool. I've just been casually watching, and got up to the 3-minute mark on my own. I thought of the first steps without watching. So, these videos are putting me on the right path to figuring out cool problems. THANK YOU.
Dont do trial and error it's bad. First apply mod 3. 615 mod 3 = 0, and 2 mod 3 = -1, however x^2 mod 3 = 1 for x relatively prime than 3. Hence y must be even. Apply y = 2k and factorize it (2^k +x)(2^k-x) = 615. This is how you prove the only integer solution is (59, 12), (-59, 12)
@@snuffeldjuret not really. Trial and error is useful in trying to solve any problems, but is never good to apply it directly. Why? Because you dont know if this trial and error pattern will continue on or not. There is not enough conclusion from just trial and error. It can gives you some valuable information but is never the way to go directly.
@@t_kon I have no idea what you mean with "apply it directly" and "the way to go directly". I am merely pointing out that "Dont do trial and error it's bad." is a blanket statement that is not always true. You can argue that it is for this task, but it is not true for all possible tasks.
*I SOLVED THIS IN MY HEAD* . I barely know how and I can't show proof, but for the sake of it I'll show my thought process; - 2^y is observably a power of 2 - Because x² should always be greater than or equal to 0, we can tell that 615 + x² should be greater than or equal to 615 - recall the sequence of the powers of 2 starting from 2^0 (1, 2, 4, 8, 16... - 32, 64, 128... - 262144, 524288, 1048576 (seriously, i know them all up until that lol) - 2^y should then be greater than 615, so we can start checking for values of x when 2^y starts at 2^10 (a.k.a 1024) 615 + x² = 1024 x² = 409 - 409 is not a square, so x isn't an integer there - Check for when y = 11 (2^11 = 2048) 615 + x² = 2048 x² = 1433 - 1433 is not a square (I guessed that, I could've been wrong lol), so x isn't an integer here either - Check for when y = 12 (2^12 = 4096) 615 + x² = 4096 x² = 3481 - It came to my head that because of the "1" in "3481", x could POSSIBLY end with the digit 1 or 9 - I go through the squares of multiples of 10 (10² = 100, 20² = 400... 40² = 1600, *50² is 2500, 60² = 3600* ) - I notice that 3481 is between 50² and 60², so therefore 50
This problem can also be solved by expressing 2^y-615 in a square form (2^(y/2)-A)^2. Here, 2^(y/2) and A must be positive integers. Otherwise, both 2^(y/2) and A are a square root of 2 multiplied with some positive integers. However, this means that x^2 is an odd power of 2 multiplied with some odd integer, which is impossible when x is an integer. From the equation A×2^(y/2+1)-A^2=615, we can check that A is a divisor of 615 such that A+615/A is a power of 2. Since 615=3×5×41 and 5+3×41=2^7, it follows that A equals either 5 or 123. In either case, we have y=12 and x=+-59.
i just solved it before watching the video by checking if sqrt(1024-615) is an integer then if sqrt(2048-615) is an integer then if sqrt(4096-615) is an integer, which it is, it’s 59.
Here's what I did. Yes, y is at least 10 for the same reason, it can't be odd for the same reason and... it can't be 18 or more: when y is even, take its square root substract 1 and square again using the typical binomial expression, and you get something smaller than 2^y-615, since 2^(y/2+1)-1 will be bigger than 615. Therefore there can't be any integer solution in those cases. For y=10,12,14,16, one could just check manually, but since x needs to be odd, the next possibility after (2^(y/2)-1) is (2^(y/2)-3) which will rule out y=14 and y=16. Then check that 2^10-615 is not a perfect square and 2^12-615 is.
In fact, we have made it a complicated one while solving it. Consider, 2^y = k. Now our equation is k -x^2 = 615. This could be written as (sqrt(k) + x)( sqrt(k) -x ) = 615. Therefore, 615 has two factors m and n (say) such that m x n = 615 where m = (sqrt(k) + x) and n = ( sqrt (k) - x). And also sqrt (k) = (m+n)/2 ans x = (m+n)/2. We have now 4 pair pairs of such values. They are (i) m = 615, n = 1 (ii) m= 205, n = 3 (iii) m = 123, n = 5 and m = 41, n = 15. Now we calculate the values of sqrt(k) and x for all these pairs of m and n as explained above. Now we have (i) sqrt(k) = (615+1)/2 = 308, x = (615-1)/2 = 307 (ii) sqrt(k) = (205+3)/2 = 104, x = (205-3)/2 = 101 (iii) sqrt(k) = (123+5)/2 = 64, x = (123-5)/2 = 59 and (iv) (41+15)/2 = 28 and x = (41-15)/2 = 18. If we observe the values of sqrt(k) in 4 alternatives, only in (iii) which is 64, it can be only expressed as a power of 2 while others are not. So this pair is the right answer. Now sqrt(k) = 64 = 2^6, so that k = 2^12. As we have assumed 2^y = k, y is now 12 and x =59.
Well if there is a single solution then no doubt primary methods are way less complicated than other methods We have, 615 + x² = 2ʸ x² must be a +ve integer => 615 + x² > 615 ∴ possible values of 2ʸ are 1024, 2048, 4096 or greater. => 2ʸ - 615 = x² 2ʸ - 615 must be a square no. (i) 1024 - 615 = 409 (not a square no.) (ii) 2048 - 615 = 1433 (not a square no.) (iii) 4096 - 615 = 3481 (is a square no. ∴ possible value of 2ʸ can be 4096 => 2ʸ = 4096 => 2ʸ = 2¹² => 𝘆 = 𝟭𝟮 And, =>x² = 3481 ....[ by (iii) ] =>x² = ∓ 59² =>𝘅 = ∓ 𝟱𝟵 Thus, x= +59 or x = -59 And y = 12 No complications..
Y has to be even. There is no other way to solve the problem. Only if this ruled out, then we can check for Y as odd. But most cases such problems will come out easily when Y is considered as even. But good insights from you to check the combination of factors. I admit that I learned something new today. Thanks.
We can also solve it by substituting y=10,11,12(2^9=512 & square of a number cannot be negative, so directly taking y=10) and then subtract 615 from it. Finally finding square root of the number obtained.
I solved it by matching using difference of squares formula. Any composite number (which has more than 2 divisors) can be presented as difference of squares, so we need to present 615 as difference in some powers of 2 and some number, so: 615 = 5x123 = (64-59)(64+59) 64^2 = 2^12 x = 59, y = 12
Who on earth came up with this awesome solution? This is more a way of thinking and a method instead of traditional math. I have learned a lot from your videos, and I have expanded the way of seeing math solutions. Especially those solutions involving geometry. Thank you very much.
Good morning! It is a intersting problem. There are some problems that folow this line, e.g., find the integers that (xy-7)^2=x^2+y^2. You can complete the square at the right side. And you get: (xy-8)^2-(x-y)^2=15. So you have to prove that y is even. If y is odd, y=2 mod3 and then x^2=2 mod3, absurd. y is even. y=2a and (2^a+x)*(2^a-x)=615 and so on...
It was easy.. 615 +x^2=2^y This can be written as: 615-2^y =-x^2 Now just thinking over this Try 2^10 i.e 1024 (but it doesn't work) Now try 2^12 i.e, 4096 You will get -3481=-x^2 By this X=59 Thus X=59 ,y=12
My solution is as follow (the main points are quite similar to the video but approach is a bit different): 615 is divisible by 3, 2^y is not, hence x^2 is not divisible by 3. If x^2 is not divisible by 3, the remainder when x^2 divide 3 can only be 1, hence 2^y divide by 3 will have remainder of 1. 2^y divide by 3 will have remainder of (-1)^y, hence y must be even. We can rewrite the equation as: 615 + x^2 = 2^(2t) -> 615 = (2^t)^2 - x^2 -> 615 = (2^t - x)(2^t+x) Since y must be positive. We just need to solve for positive x as well -> 615 > 2^t + x -> t y 615 -> y >= 10. We have y is an even number from 10 to 18. Try all value and we find y=12 and x=59 or -59
One of the most enjoyable problems I have seen for quite a while . Finding the solution as explained by Presh is not easy but it *is* very satisfying . Me, I simply used the trial and error method! Noting that 2^y had the be *at least* 2^10, I only had to try 1024 minus 615 , 2048 minus 615 and 4096 minus 615 to find which of these three differences gives a perfect square . Not a whole lot of work !
Set z=y/2. Then 615=(2^z+x)(2^z-x). Factor 615 and you quickly find z=6 and x=59 or -59. I do not initially assume z is an integer but the final result is an integer as hoped for
If you move X^2 to the right and assume y=even, break it into (x+2^(y/2))(2^(y/2)-x). The prime factors of 615 are 1 3 5 and 41. 2^6 is 64. 41*3=123, 123-64=59, 64-5 is also 59, meaning we've found at least one answer, y=12 and x=+-59
I found the solution in under 2 minutes in Excel. I calculated the left hand side for a range of 1 to 75 and the same for the right. If x = 59 and Y=12 the result is 4096. Sometimes brute force is quicker and easier.
x=59 or -59 and y=12. Hit and trial. Cause 615 + x^2 can only be a perfect square of 2 so we are left with 1024, 2048, 4096, 8192 and so on. It’s hits in the third try in 4096 and we get our answer. Easy peasy within 2 minutes anyone can do it. Edit: I know this won’t work every time and is not efficient but sometimes you can see an easy option in every problem. Even though you couldn’t find it, it would just take a minute or two and you can compensate it every now and then. I personally find this very helpful in tests and exams as I am not very good at maths but I am good at finding alternate ways to the problem. Also sorry for my bad english.
I provide an alternative solution which I think it can easily to understand. 2^y is even number, so that x is odd number. y cannot be an odd number because 615+x^2 (left hand side) can be a prefect square but 2^y (right hand side) cannot when y = odd number integers. 615+x^2 can be rewrite as 1 + 2*(1)(307) + x^2, when x = 307, 615+x^2 = 308^2 is a prefect square while 2^y cannot be a prefect square when y= odd number integers. Contradiction occurs. So that y is even number. Rewrite the question as 2^y-x^2=615, (2^(y/2)+x)(2^(y/2)-x)=615. Since y is even number, 2^(y/2) is integer, so that we need to factorised 615. 615=3*5*41. Let 2^(y/2)+x = a, 2^(y/2)-x = b, then a + b = 2^((y/2)+1) is index of 2, a-b = 2x The solution for a and b may be a = 3, b = 205 or a = 205, b = 3, => a + b = 208, is not index of 2. a = 5, b = 123 or a = 123, b = 5, => a + b = 128, is index of 2. a = 15, b = 41, or a = 41, b = 15. => a + b = 56, is not index of 2. It is not difficult to see that only a = 5, b = 123 or a = 123, b = 5 can satisfy the equation. x= (a-b)/2 => x = -59 or x = 59. a + b = 2^((y/2)+1) = 128 => y/2 +1 = 7, y =12.
Apologies for the re-upload. In the previous version I only solved for positive integers. It is straightforward to then find the other solution. But I uploaded this corrected video presenting both because not everyone reads the comments. Also these videos are getting shared in a lot of places so I'm doing my best to present accurate mathematical results. I thank Aniket Gupta and Ryan Wilson for pointing out the oversight!
What is your email?
So where did I go wrong in the original video? I often get emails where people ask me to look for mistakes in their work, if they did not find the correct solution in one of my videos. One important skill in mathematics is reviewing your own work. It's not always fun, but it's really important to learn where you might make errors. So where did I go wrong? In my self-review, there are three places I could have realized my mistake. Place #1: I said the difference of factors is (2^n + x) - (2^n - x) = 2x. Then I solved 2x = 118 to get x = 59. The thing is the difference of factors can also be (2^n - x) - (2^n + x) = -2x. Setting that equal to 118 would have given x = -59. Place #2: When I solved y = 12, I could have then solved 615 + x^2 = 2^12 where I could have seen x = 59 or -59. Place #3: In checking my work, I should have seen there was an x^2 term, and it's a common mistake to overlook a negative solution. This is the kind of self-criticism I do for my math homework. I have noticed people don't like to admit if they are wrong. But I have found self-reviews and honesty are big strengths to improving math skills. So I thank all the legitimate criticism (if accompanied with legitimate honesty) for these videos--you're helping us all get better at math.
@@MindYourDecisions your email please. I have an important question that no one in Nepal has solved.
Pranesh Pyara Shrestha his email is at every end of his video
@@Lqtan16 oh I found it
Can you imagine solving all the question and forgetting to include -59 to the equation get 0 because of it.
That's sooooo sad
Lol i did same😂
@@Vegan_PhysicsEnthusiast u definitely dont have these questions
@@namanlakhotia6393 if you do prmo training you will definitely get this type of questions
@@funni111 prmo is not comp in schools though
literally a week after every video criticizing wolfram alpha's capabilities is posted, they end up fixing it
www.wolframalpha.com/input/?i=solve+615+%2B+x%5E2+%3D+2%5Ey+over+the+integers
@@ashyourresidentenby5916 solve 604 + x^3 = 3^y over the integers lmao still cant solve these types of questions
@@tgdhsuk3589 x is 5 and y is 6 HA solved it in like 15 seconds
@@tgdhsuk3589 WolframAlpha solves this
Ethical hacking..
*99% of the people who clicked the video underestimated this problem*
@BlazePlayz YT of course you can just blindly type in numbers, that´s easy, but what if the answer was smth like y=22
i think you would have some troubles then
@BlazePlayz YT Yep the key is to spot 3481 as a square
Rishi Jai is there a certain trick to spotting a high number with a square root?
Cus I did the blindly squaring numbers on my calculator around the sums + 615 being a power of 2 thing...
I did but I solved it too
@TH-cam Watcher i also did the same😁
7:27 "All we did was use of principles of number theories."
Right, very simple...
WYSI
He:this is presh talwalkar
Captions:this is fresh lakewater
pressure walker
dash fallwalker
it says Presh TalWalker here
It says pressure locker
Not Luke Skywalker.
It is easier to solve this problem with little logical way than to follow Presh`s complicated method. Look at the problem once. 2^y is always an even number irrespective of whether y is even or odd. Now we can write the equation as 615 =2^y - x^2. Here 615 is an odd number. Therefore, x^2 will have to be an odd number. Because, the difference between two even numbers is always even and the difference between an even number and an odd number is always odd. Since x^2 is an odd number, x must be an odd number too as the square of an odd number is always an odd number and the vice-versa is also true. Let x = 2n+1 for any value of n, odd or even. Now we have 2^y - (2n+1)^2 = 615. This can be written as (2^(y/2))^2 - (2n+1)^2 = 615. This now becomes the difference between two squares form. That means the left hand side will be the product of [(2^(y/2)) + (2n+1)][2^(y/2) - (2n+1)]. Putting, 2^(y/2) = a, and 2n+1 = b for simplification, we have (a+b)(a-b) = 615. Now 615 has to be the product of two numbers. The number 615 = 1 x 3 x 5 x 41. Therefore, we have four alternative pair of numbers whose product will be 615. They are (i) a+b = 615 and a - b = 1 or (ii) a + b = 205 and a - b = 3 or (iii) a + b =123 and a - b = 5 or (iv) a + b = 41 and a - b = 15. Solving (i) we have a = 308, b = 307 (ii) a =104, b = 101 (iii) a = 64, b = 59 and (iv) a = 28, b = 13. Here note that 'a' = 2^(y/2) that is a number power of 2. If we check the values of 'a' in all thee alternatives, only (iii) which is 64,satisfies our condition and the others are not.Therefore 2^(y/2) = 64 = 2^6. That is y/2 =6, That is y = 12 and b = 2n+1 = x =59.
You assumed a to be an integer without proving that y is even.
@@xmarteo I have simplified it in a better way in my next explanation. Pl look at it again.
@@rcnayak_58 you don't even have to set b = 2n+1 because you don't use it again.
Also, xmarteo's comment raises a valid concern, because then you would have to check for more divisors (divisors on Z[sqrt(2)] ). Or you could show that y is even, resulting in the same proof as MindYourDecisions.
Ultimately, your proof is not shorter than MindYourDecisions'.
@@rcnayak_58 Try your logic with 23 + x^2 = 2^y. You'll get that (a+b)(a-b) = 23, which only has one integer solution: (a, b) = (12, 11), and since 12 is not a power of 2, you would conclude that there's no solution with integer x and y. But (x, y) = (3, 5) is a solution, corresponding to (a, b) = (4sqrt(2), 3).
@@xmarteo 2^y is always even because it's a multiple of 2 and an even number is defined as multiple to, so by definition 2^y is even. The problem though, is that this solution is almost as complex as Presh's, because it practically requires the same methodology of thinking. While I agree it might be efficient, you've to keep in mind that he was send this solution as well.
UPDATE : Wolfram Alpha can now solve this problem. I loved your approach.
You can solve it beautifully in Excel by solving (with parameters) an equation as 615+x^2 - 2^y = 0
Alternatively you can calculate y=log(615+x^2)/log(2) and then an integer value of y should be found. All to be solved in Excel :)
I'd give you a cookie, but you only really deserve it if you use SQL. (By the way, good luck brute-forcing the proof that there are no other solutions.)
After spotting that n > 9, I just used trial and error with n = 10, 11... to check which values would give a perfect square for 2^n - 615. Eventhough, trial and error isnt the 'best' way to solve problems, I think this would be a quicker approach :)
THOUGH I ABSOLUTELY LOVED THE WAY YOU DID IT, IT IS ALWAYS FUN TO KNOW HOW THE OTHER PERSON SOLVED IT. I LOVE MULTIPLE APPROACHES
Hey presh. Just wanted to thank you... we had a really difficult math test that was more about thinking than calculating. It was really hard but everytime i started a new question, I thought about what Presh will do and how he will solve that, and that helped solving most of the problems. Thanks for improving and sharpening the thinking for me, and for other hundreds of thousands people
ethan frommer ok boomer
@@sahaj9810 wait how did u get the 2^10 , 2^11, and 2^12 are the only possibilities? im a little confused.
@@sahaj9810 he wanted to say what about 2¹³ .. 2¹⁴..2^15 and so on
Way simpler: x=sqrt(2^y-615)
So 2^y has to be > 615
By trial and error starting from 1024 we can quickly reach x an integer (59) at 4096.
Ye prove it that that's the only answer
I was with you right up to "You're watching Mind Your Decisions"....After that I said, "Huh?"
Fresh tall walker
@@SameerKhan-fd2qe Fresh Tail Wagger
I feel proud for solving this problem not only alone but also with different thinking :))
Can you share your way?
@@kwcy92
Of course! It was a guessing game with math background 😂😂😂
We can rewrite the problem as 615=2^y-x^2
I know x will be an odd number (because 615 is odd and 2^y is always even). Because we know powers of two, we want a “bigger” number for y because difference needs to be 615
From there on I played a bit with numbers to see what happens when I use bigger x, smaller y... I started with x=y=9 and moved up by 5-10 for x and one for y and got there relatively fast. I know it’s not the best way to solve, but it worked 😂😂😂
615 is divisible by 3, hence 2^y-x^2 is divisible by 3. Easily checking out we could see that 2-x^2 is not divisible by 3, then 2^y-0 or 2^y-1 is divisible by 3. Since 2^y-0 isn't, we can conclude that 2^y-1 is, then y is even
That's interesting!
I tried taking the log base 2 on both sides in order to isolate y.
When I first started to solve it, i gave up. But then I decided to try it again and got it, using (x-1)^2 = x^2 - (2x-1)
Am I the only one who always gets depressed when he says:'Did you figure it out'? . 'cause I am already happy when I can follow his explanation.
Yep, I’m with you. Don’t usually even understand the explanation it seems.
the first thing i did when i saw this problem was google ‘what does solve over the integers mean?’. i promise you’re not alone.
Wolfram & Hart couldn’t solve it, but Winifred Burkle could!
Thanks for this genius solution.
I also started as Nayak's solution: x has to be odd, then x=2n+1. Then, we have that 4n(n+1)=2^y-616, or n(n+1)=2^(y-2)-154. Now I went with y-2>=8, and found y-2=10 and n=29, which give x=59 and y=12. Cheers, Marco.
I always depend on Wolfram Alpha for solving Equations in Physics. But this new information blows my mind. Now I have to go back and do all those tedious exercises from Goldstein.
It works now 🧐
For mathematics, only use wolfram for getting hints toward a solution. Often wolfram does not give you all possible solutions or even managed to find them (this has happened to me a lot in calculus and even combinatorics). Try to understand the problem and the theory behind it and solve it from there. There are lots of mathematics that are not implemented yet in wolfram either.
I thought it was an algebra problem. Then I kept watching and my brain melted.
Same
@mindyourdecisions I took a game theory class in college this past semester (spring 2018) and you would not believe the amount of times you were listed as a source.
Great and imaginative solution!
Lots of people commenting on these problems don’t understand mathematics though. They believe one only has to do trial and error to find one solution, but forget all about finding every possible solution while also showing that there can be no other. This is the core thinking that should be taught. Wonder what they actually teach people in school nowadays.
The solution to this problem really blowed my mind. The structure of it is the most imaginative and trickery of what I've encountered so far 😱
Thanks for sharing
It's easy, wolframalpha cannot make a sandwitch, but I can!
What, make a witch out of sand? Well, it is nearly Halloween, after all. So I guess all you need is to go to the beach!!
Fred
but can you tho?
Dont provoke them wolfram alpha is listening
Did you try "sudo make a sandwich"? That usually does the trick
In this video, he looks mod 10 to get that y is even. But it is much easier to look mod 4 to get that fact.
And also, a lot of people are saying they solved this by guessing and checking. While that does find a solution, it does not find a proof of the only solutions, which this video does.
Nope. The video didn't prove that this is the only solution
@Bragadeesh S yes it did
My bad. Agreed
I'll see your mod 4 and raise you mod 3. In fact, modulo 3 you get 615 ≡ 0, x² ≡ 0 or 1 and 2^y ≡ (-1)^y, so y must be even.
Neat. I used the mod 10 solution myself, which seemed more obvious to me since I count in base 10 and tend to notice base-10 patterns before considering other bases. (OK, I lie. I did briefly look at mod 2 first, but that only got me as far as "x must be odd" before I abandoned that approach and jumped to base 10. Mod 3 and 4 never occurred to me.)
2^y - 615 must be a square. Went on substituting y > 9
That's cool.
One more analysis, that 2^y is always even so sum of of 615 and x^2 should also be even, which is possible when x^2 is odd.
x^2 is odd when x is odd integer.
Hence odd integral solutions of X is only possible.
I also wwnt on substituting y value but soon after I though that this will not help me for IIT JEE
I am in 10
Going go attempt jee in 2022
But i got the answer by putting 12
I did the same thing. Once he started looking at digit repetition, I was like "y tho?"
Take the logarithm base 2 of both sides:
Answer: y = log(x^2 + 615)/log(2) + ((2 i) π n)/log(2) for n element Z
x = ± 59, y = 12
(that's a Wolfram's answer)
This is the first thing I thought, surprised I had to go so low to find this comment
I‘ve been taking some number theory classes in my free time and we‘ve just started with modular arithmetics, and I was so happy when I got the solution before watching the video haha :) I did it mod 3 tho
Same here :D
I solved this in about 60 seconds from the thumbnail. The right hand side has to be 1024 ... 2048 ... 4096 etc. So just keep trying until you find one which has an integer square root. Bingo - square root of 4096 - 615 is 59 (and -59). A familiarity with well known binary numbers helps. Because I solved it before I opened the video the no calculator rule didn’t apply. 🙂
Cool bean
Holy crap the captions got your name right. Well, there's a space between tal and walker but damn, I'm glad I've been checking every video haha
As always with these videos, a lot of people don't grasp the concept of PROOF.
Well how do you say? There is no any proper explanation about proof. It's so abstract and we never realized we need to use inequalities for this kind of problem. You know in HS we always use regular method, but not with this long and tiring explanation only to find y=2n. How come common people realized that? Me as a Master degree student even cannot realize this
nonesuch27 Most people don't grasp the concept of proof. Proof is really only needed in the study of pure mathematics. For everyday applications of mathematics, you don't need proofs, just like you don't need a degree in automotive engineering to be an auto mechanic.
What? What does proof have to do with anything? The problem as worded says nothing about proof. It asks you to find x and y, only allowing integers. Period.
@@qc1okay If it asks to SOLVE the problem then it means you have to find ALL the solutions. If you find only ONE solution, you have to prove its the only solution. Otherwise you didn't solve the problem. Many people just say "easy, i guesed it in 3 tries", while that worked in this case because there's only one solution, it won't always work.
@@dariobarisic3502 No it doesn't. Just that simple. Solve does not mean all. This TH-cam channel isn't a formal math class, and even if it were, its instructions must be clear to its viewers. If "all" were wanted, "all" had to be stated. Incidentally, I went thru the full process Presh used, and it both leaves out steps and overcomplicates others.
Here's how I solved this with paper and pencil. Rewrote expression as x^2 -2^y=-615. X^2 will always be a positive integer value, so 2^y must be some power of 2 greater than 615 where the difference between the two expressions has a (positive) integer square root. 2^10 = 1024 & square root of the difference (409) is not an integer, 2^11=2048 and square root of the difference (1433) is not an integer, 2^12=4096, subtracting 615 yields 3481. Since the last two digits of 3481 are 81, there is a good chance that 59 could be the square root. I tried that and it worked. Thus, x=59, y=12.Yes, I know intuition and trail and error aren't proper 'solutions.' Thanks for this channel!
I solved this in about 3 minutes by simply listing the powers of two, subtracting 615, and looking for an integer square root. The first one even possible had to be more than 615, so I only had to check three powers of two before finding one with an integer square root.
2^y , which means it could only be2,4,8,16,32...
Therefore,just to find out which 2^y number can be square rooted after minus 615.There is 615, so we can start with 1024,2048,4096... then find out that the answer is x=59,y=12
Thats a numeric solution, not analitic. Prove that that is the only solution.
I did trial and error method and solved it in 2mins.....2^9=512 & 2^10=1024, 2^y has toh be greater than 615, so consider 2^10 =1024, so 1024-615=409 which is not a Perfect Square, 2^11=2048, so 2048-615=1433 which is also not a Perfect Square, and then 2^12=4096, so 4096-615=3481 which is Square of 59. Thus x=59 and y=12
Rahul Chhabaria sameee hahaha i tried 1024,2048, then finally 4096 and it worked!!
-59 is second matching X :D
This is what I did. No need to make it more complicated than it is :P
@@ceegers Actually, it is needed to prove that no other solutions exist
@@Cohnan13 Some people here don't what "solving" means
Brilliant! This is the content I can't get enough of.
I just did guess and check and found 59 as a solution for x. Obviously 2^y has to be greater than 615, so I started with 1024-615, which wasn't a perfect square, so I did 2048-615, then 4096-615. I found the square root to be 59. So -59 and 59 are solutions when y=12. I imagine there are infinite solutions.
EDIT: The solution in the video is much more elegant, and also shows there is only one solution. Very cool.
I have to admit, when I first saw this problem I completely did not think it would be very complicated. Then I tried it and got stuck but I was still convinced there some basic algebra trick that I'm missing. Then after watching this solution, I realize there was no chance I was solving this without the explanation
Literaly just had this in a test yesterday and couldnt answer, cmon universe
Not possible
Be honest and confess; how many of you guys checked whether wolfram alpha could solve this or not? Well, I did
and can it?
Nope
it actually gives you integer solutions if you type just the equation in it.
I don’t see why not, and it was able to
th-cam.com/video/eEyQyixOGpU/w-d-xo.html
Wolframalpha: isolate y on 615+x^2=2^y solves the problem
I solved it partly using logic and finished it using trial and error, but your method is so much better because it demonstrates the logic required to prove it is the only solution. I look forward to sharing this with my senior students.
Sometimes you come across a map problem and you think “I get it” and you “solve” the problem under a couple of minutes only to realize later the problem was above your “pay grade” and a much more complicated problem that you anticipated. This has happened to me back to back today ;)
3:52 it was even easier to establish this by looking at the equation mod 3
That's bit advances.
Presh Talwakar makes things intuitive.
@@TechToppers not realy that hard. Every square is 0 or 1 mod 3, every even power of 2 is 1 mod 3, but every odd power of 2 is 2 mod 3. So the y is even. You would need to prove the initial statements though (but it's very simple).
@@d4slaimless I know it is simple. But if you can avoid technicalities, then why not!
@@TechToppers I am not arguing that the solution in the video is very demonstrative. Just saying that going for the reminder of a division is not that hard either.
wow i found a cuber under a maths video!
Who else used a different strategy, but found a much faster way to do it? My strategy was to try different powers of 2 (greater than 9). For example: 615 + x^2 = 2^11 (2048). This implies that x^2 = 1433. That isn't a square value (for a positive even integer), so I tried 615 + x^2 = 2^12 (4096). That meant x^2 was equal to 3481. Since 3481 is a square, I was able to solve for x and y.
Wish you could see my raised hand ✋ ✋
Ron Madan but you never proved or showed that was the only solution. That is the challenge with these problems.
Try or not try, that is not the question. Use the force of logic, Luke!
This was a very fine problem! It was good that 615 is quite easy to divide into prime factors, otherwise it wouldn't have been so easy without a calculator :)
The approach of checking the last digits, as all simple thinking, is genius!!
Great way to solve such a problem.
I solved this directly by the substitution 2 to the yth power = a squared, which gets to the answer more quickly, although the remainder of the approach is similar to what was presented.
The above approach leads immediately to (a-x)(a+x) = 615 and by substituting the four ways to factor 615 (41x 15, 123 x 5, 205 x 3, and 615 x 1), the only pair of factors that produce integer solutions is 5 and 123. a - x = 5 and a + x = 123 leads to a = 64 and x = 59, and since a squared = 2 to the yth power, 2 to the yth power must equal 4096, so y = 12.
I solved the equation in seconds using a chart containing square of numbers upto 100
It was easier than the method in the video
Why did WolframAlfha give up when it reach x = 3?
AI won't take over the world so soon.
@Presh, had you constructed the proof the other way around, it would have been a lot closer to how people/ students think. Start with observing x^2 and that if y is even then you can do a^2 - b^2 = 615 which is easy to solve. Then prove that y cannot be odd, thus the only solutions are the ones where y is even (x = +/- 59, y = 12).
The way you started it by observing y has to be even and then stating ("This will be a key in our finding the solution") makes you look insightful, but does not help people develop thinking patterns in math problem solving (especially Diophantine equations)
Well... I solved the problem the exact same way and in the exact same order. It's not really about being insightful but about using the fact integers are a strong constraint on the solutions.
@Dan It's the same with last digit analysis. How do you know it is we need in this specific question? You don't know, but i bet factoring leads to results more often than last digit analysis.
If you look at the equation in mod 3 it is clear that y is even which means that you can move x^2 to right side of the equation and do difference of two squares. And then try for positive divisors of 615
I just started at 1024 and subtracted 615 from each power of 2 from there on. 3rd try, 4096. Obviously if it was something like 2^2000, that would take far too long, but for this one...worked in like 20 seconds.
1024 - 615 = 409 (not a square)
2048 - 615 = 1433 (not a square)
4096 - 615 = 3481 (59^2)
I think this is why I hate traditional math problems, they make them simple enough that you can brute force an answer instead of doing the work. Got me through all my trig/calc classes in high school :D
I did the same in about 20 seconds, too!
You can also find that y is even using mod 3. A little faster than mod 10
Tom Domenge sure, (mod 3) 615 = 0, x^2 = 1 or 0, and 2^y = (-1)^y. So 615 + x^2 = 0 or 1. If y is odd, 2^y=(-1)^y = -1 which is impossible. So y is even
Explain the Difference between Quasi-Static and Non Quasi-Static
@Mindyourdecisions
1) This isn't a physics channel
2) @s don't work anymore
Minha solução:
Se "==" representa congruência, temos:
615+x²==x² (mod 3).
2ⁿ=/=0 (mod 3)
→x²=/=0 (mod 3)
→x²==1 (mod 3).
2ⁿ==1 (mod 3)
→n=2a
615=2²ª-x²→
615=(2ª+x)(2ª-x).
Edit: ok, vamos testar todos os pouquíssimos casos: 615=3•5•41, além disso, 2ª+x>2ª-x. As únicas possibilidades são (supondo x≥0):
2ª+x=615 e 2ª-x=1→2•2ª=616. Abs
2ª+x=205, 2ª-x=3→2•2ª=208. Abs
2ª+x=41 e 2ª-x=15→ 2•2ª=56. Abs
E
2ª+x=123 e 2ª-x=5
2ª=64, de onde segue que a única
solução é a=6→n=y=12 e x=±59
parabens cara
Translation for english:
My solution is:
If "==" represents congruence then we have:
....
Testing all the possible cases (which are just a few) , we have that the only solution is...
@@Gutagi macho, são bem poucos casos
SilverBlade como assim ???
@@Gutagi taí. Resolvido completo (mas é tão trivial que eu nem devia ter perdido meu tempo editando)
X^2=2^y-615. For x^2 to be positive, y is at minimum 9. Then increase y incrementally until 2^y-615 is the square of an integer.
To chekc whether y is even or odd we can simply use 3 as x^2 gives remainder 0 or 1 when diveded by 3 and 2^y gives remainder 2 when y is odd and 1 when is even therefore y must be even
Your videos are awesome
@@iotaop9573 always always
hello please look into integrals by the Feynman technique
Thanks macOS for "Grapher"
For the ones using a computer or a calculator, once you observe that y has to be even, therefore 2^y is a square.
Notice that once you're at a perfect square n², you can only get to another square by substracting 615 only as long as n is small? How small you ask,
(n+1) ² - n² > 615, this gives n>307.
Therefore once 2^y exceeds 307² , there's no point checking.
Therefore you only need to check for the values of y= 10,12,14,16
This is so cool. I've just been casually watching, and got up to the 3-minute mark on my own. I thought of the first steps without watching. So, these videos are putting me on the right path to figuring out cool problems. THANK YOU.
Dont do trial and error it's bad.
First apply mod 3. 615 mod 3 = 0, and 2 mod 3 = -1, however x^2 mod 3 = 1 for x relatively prime than 3. Hence y must be even. Apply y = 2k and factorize it (2^k +x)(2^k-x) = 615. This is how you prove the only integer solution is (59, 12), (-59, 12)
trial and error isn't bad, it is an extremely valuable tool in figuring things out. When applying math on the real world, trial and error is crucial.
@@snuffeldjuret not really. Trial and error is useful in trying to solve any problems, but is never good to apply it directly. Why? Because you dont know if this trial and error pattern will continue on or not. There is not enough conclusion from just trial and error. It can gives you some valuable information but is never the way to go directly.
@@t_kon I have no idea what you mean with "apply it directly" and "the way to go directly". I am merely pointing out that "Dont do trial and error it's bad." is a blanket statement that is not always true. You can argue that it is for this task, but it is not true for all possible tasks.
Is doing mod 3 also trial and error ?
Take mod 16 on both sides we end with x = +3 or -3 mod 16 .But 59 is neither .Where am i going wrong.
I did it completely differentlymin my head in a third the amount of time it took to explain that solution
r/humblebrag
Let me guess... you guessed it and now think you got a a proof.
Don’t use calculator but count 2^9 and factor out 615 :D
That wasn't that hard, was it?
You need a calculator for the first few powers of two? Like really?
The worst students know 2 powers from 0 to 20
@@AvelinoTiago No.
*I SOLVED THIS IN MY HEAD* . I barely know how and I can't show proof, but for the sake of it I'll show my thought process;
- 2^y is observably a power of 2
- Because x² should always be greater than or equal to 0, we can tell that 615 + x² should be greater than or equal to 615
- recall the sequence of the powers of 2 starting from 2^0 (1, 2, 4, 8, 16...
- 32, 64, 128...
- 262144, 524288, 1048576 (seriously, i know them all up until that lol)
- 2^y should then be greater than 615, so we can start checking for values of x when 2^y starts at 2^10 (a.k.a 1024)
615 + x² = 1024
x² = 409
- 409 is not a square, so x isn't an integer there
- Check for when y = 11 (2^11 = 2048)
615 + x² = 2048
x² = 1433
- 1433 is not a square (I guessed that, I could've been wrong lol), so x isn't an integer here either
- Check for when y = 12 (2^12 = 4096)
615 + x² = 4096
x² = 3481
- It came to my head that because of the "1" in "3481", x could POSSIBLY end with the digit 1 or 9
- I go through the squares of multiples of 10 (10² = 100, 20² = 400... 40² = 1600, *50² is 2500, 60² = 3600* )
- I notice that 3481 is between 50² and 60², so therefore 50
This solution is so brilliant, I'm stunned...
7:36 so now you say this every time. ok.
WolframAlpha solve it for now
If you do trial and improve it's easier I got it in less than 5 minutes
Thing is finding a solution by trial and error doesn't prove there's no other solution
This problem can also be solved by expressing 2^y-615 in a square form (2^(y/2)-A)^2.
Here, 2^(y/2) and A must be positive integers. Otherwise, both 2^(y/2) and A are a square root of 2 multiplied with some positive integers. However, this means that x^2 is an odd power of 2 multiplied with some odd integer, which is impossible when x is an integer.
From the equation A×2^(y/2+1)-A^2=615, we can check that A is a divisor of 615 such that A+615/A is a power of 2. Since 615=3×5×41 and 5+3×41=2^7, it follows that A equals either 5 or 123. In either case, we have y=12 and x=+-59.
i just solved it before watching the video by checking if sqrt(1024-615) is an integer then if sqrt(2048-615) is an integer then if sqrt(4096-615) is an integer, which it is, it’s 59.
Apply simultaenous equation and it's done in a second.
How??
I solved for x
X=√2y−615
Im gonna be honest with yall. I didnt managed to do it.
Here's what I did. Yes, y is at least 10 for the same reason, it can't be odd for the same reason and... it can't be 18 or more: when y is even, take its square root substract 1 and square again using the typical binomial expression, and you get something smaller than 2^y-615, since 2^(y/2+1)-1 will be bigger than 615. Therefore there can't be any integer solution in those cases. For y=10,12,14,16, one could just check manually, but since x needs to be odd, the next possibility after (2^(y/2)-1) is (2^(y/2)-3) which will rule out y=14 and y=16. Then check that 2^10-615 is not a perfect square and 2^12-615 is.
In fact, we have made it a complicated one while solving it. Consider, 2^y = k. Now our equation is k -x^2 = 615. This could be written as (sqrt(k) + x)( sqrt(k) -x ) = 615. Therefore, 615 has two factors m and n (say) such that m x n = 615 where m = (sqrt(k) + x) and n = ( sqrt (k) - x). And also sqrt (k) = (m+n)/2 ans x = (m+n)/2. We have now 4 pair pairs of such values. They are (i) m = 615, n = 1 (ii) m= 205, n = 3 (iii) m = 123, n = 5 and m = 41, n = 15. Now we calculate the values of sqrt(k) and x for all these pairs of m and n as explained above. Now we have (i) sqrt(k) = (615+1)/2 = 308, x = (615-1)/2 = 307 (ii) sqrt(k) = (205+3)/2 = 104, x = (205-3)/2 = 101 (iii) sqrt(k) = (123+5)/2 = 64, x = (123-5)/2 = 59 and (iv) (41+15)/2 = 28 and x = (41-15)/2 = 18. If we observe the values of sqrt(k) in 4 alternatives, only in (iii) which is 64, it can be only expressed as a power of 2 while others are not. So this pair is the right answer. Now sqrt(k) = 64 = 2^6, so that k = 2^12. As we have assumed 2^y = k, y is now 12 and x =59.
Really very imaginary method....
Interesting....
imaginative? (i'm not native english speaker)
Solved it in 1 minute...by substitution method!!!😊😅😅
Same
Bro there wasn't 2 equations, how did you solve it plz tell
If a scientist couldnt solve it,
Why would I
All competent scientists can solve that. It's easy.
Well if there is a single solution then no doubt primary methods are way less complicated than other methods
We have,
615 + x² = 2ʸ
x² must be a +ve integer
=> 615 + x² > 615
∴ possible values of 2ʸ are
1024, 2048, 4096 or greater.
=> 2ʸ - 615 = x²
2ʸ - 615 must be a square no.
(i) 1024 - 615 = 409 (not a square no.)
(ii) 2048 - 615 = 1433 (not a square no.)
(iii) 4096 - 615 = 3481 (is a square no.
∴ possible value of 2ʸ can be 4096
=> 2ʸ = 4096
=> 2ʸ = 2¹²
=> 𝘆 = 𝟭𝟮
And,
=>x² = 3481 ....[ by (iii) ]
=>x² = ∓ 59²
=>𝘅 = ∓ 𝟱𝟵
Thus,
x= +59 or x = -59
And y = 12
No complications..
Y has to be even. There is no other way to solve the problem. Only if this ruled out, then we can check for Y as odd. But most cases such problems will come out easily when Y is considered as even. But good insights from you to check the combination of factors. I admit that I learned something new today. Thanks.
We can also solve it by substituting y=10,11,12(2^9=512 & square of a number cannot be negative, so directly taking y=10) and then subtract 615 from it. Finally finding square root of the number obtained.
I solved it by matching using difference of squares formula. Any composite number (which has more than 2 divisors) can be presented as difference of squares, so we need to present 615 as difference in some powers of 2 and some number, so:
615 = 5x123 = (64-59)(64+59)
64^2 = 2^12
x = 59, y = 12
Who on earth came up with this awesome solution? This is more a way of thinking and a method instead of traditional math. I have learned a lot from your videos, and I have expanded the way of seeing math solutions. Especially those solutions involving geometry. Thank you very much.
It is traditional math, a branch called number theory.
Good morning! It is a intersting problem. There are some problems that folow this line, e.g., find the integers that (xy-7)^2=x^2+y^2. You can complete the square at the right side. And you get: (xy-8)^2-(x-y)^2=15.
So you have to prove that y is even. If y is odd, y=2 mod3 and then x^2=2 mod3, absurd. y is even.
y=2a and (2^a+x)*(2^a-x)=615 and so on...
It was easy..
615 +x^2=2^y
This can be written as:
615-2^y =-x^2
Now just thinking over this
Try 2^10 i.e 1024 (but it doesn't work)
Now try 2^12 i.e, 4096
You will get
-3481=-x^2
By this X=59
Thus X=59 ,y=12
i=10
y=-1
x=-1
while (True):
a = 2**i - 615
if (int(a**.5)*int(a**.5)==a):
y =i
x= int(a**.5)
break
i+=1
print(x,y)
this is the solution in python
My solution is as follow (the main points are quite similar to the video but approach is a bit different): 615 is divisible by 3, 2^y is not, hence x^2 is not divisible by 3. If x^2 is not divisible by 3, the remainder when x^2 divide 3 can only be 1, hence 2^y divide by 3 will have remainder of 1. 2^y divide by 3 will have remainder of (-1)^y, hence y must be even. We can rewrite the equation as:
615 + x^2 = 2^(2t) -> 615 = (2^t)^2 - x^2 -> 615 = (2^t - x)(2^t+x)
Since y must be positive. We just need to solve for positive x as well
-> 615 > 2^t + x -> t y 615 -> y >= 10. We have y is an even number from 10 to 18. Try all value and we find y=12 and x=59 or -59
We can eliminate all values of y
One of the most enjoyable problems I have seen for quite a while . Finding the solution as explained by Presh is not easy but it *is* very satisfying . Me, I simply used the trial and error method! Noting that 2^y had the be *at least* 2^10, I only had to try 1024 minus 615 , 2048 minus 615 and 4096 minus 615 to find which of these three differences gives a perfect square . Not a whole lot of work !
Me too.
Set z=y/2. Then 615=(2^z+x)(2^z-x). Factor 615 and you quickly find z=6 and x=59 or -59. I do not initially assume z is an integer but the final result is an integer as hoped for
If you move X^2 to the right and assume y=even, break it into (x+2^(y/2))(2^(y/2)-x). The prime factors of 615 are 1 3 5 and 41. 2^6 is 64. 41*3=123, 123-64=59, 64-5 is also 59, meaning we've found at least one answer, y=12 and x=+-59
I found the solution in under 2 minutes in Excel. I calculated the left hand side for a range of 1 to 75 and the same for the right. If x = 59 and Y=12 the result is 4096. Sometimes brute force is quicker and easier.
x=59 or -59 and y=12. Hit and trial. Cause 615 + x^2 can only be a perfect square of 2 so we are left with 1024, 2048, 4096, 8192 and so on. It’s hits in the third try in 4096 and we get our answer. Easy peasy within 2 minutes anyone can do it.
Edit: I know this won’t work every time and is not efficient but sometimes you can see an easy option in every problem. Even though you couldn’t find it, it would just take a minute or two and you can compensate it every now and then. I personally find this very helpful in tests and exams as I am not very good at maths but I am good at finding alternate ways to the problem. Also sorry for my bad english.
Mind your decision is best Chanel for students
I provide an alternative solution which I think it can easily to understand.
2^y is even number, so that x is odd number. y cannot be an odd number because 615+x^2 (left hand side) can be a prefect square but 2^y (right hand side) cannot when y = odd number integers.
615+x^2 can be rewrite as 1 + 2*(1)(307) + x^2, when x = 307, 615+x^2 = 308^2 is a prefect square while 2^y cannot be a prefect square when y= odd number integers. Contradiction occurs.
So that y is even number. Rewrite the question as 2^y-x^2=615, (2^(y/2)+x)(2^(y/2)-x)=615. Since y is even number, 2^(y/2) is integer, so that we need to factorised 615. 615=3*5*41.
Let 2^(y/2)+x = a, 2^(y/2)-x = b, then a + b = 2^((y/2)+1) is index of 2, a-b = 2x
The solution for a and b may be
a = 3, b = 205 or a = 205, b = 3, => a + b = 208, is not index of 2.
a = 5, b = 123 or a = 123, b = 5, => a + b = 128, is index of 2.
a = 15, b = 41, or a = 41, b = 15. => a + b = 56, is not index of 2.
It is not difficult to see that only a = 5, b = 123 or a = 123, b = 5 can satisfy the equation.
x= (a-b)/2 => x = -59 or x = 59.
a + b = 2^((y/2)+1) = 128 => y/2 +1 = 7, y =12.