@@allasar For much larger numbers the way BeeBee does it will be faster - take the number 734,413. The next perfect square that is bigger than this number is 857*857 = 734,449, Notice that 734,449 - 734,413 = 36 which is also a perfect square and so the factors of my number are 857-6 and 857+6 = 851 and 863. 851 * 863 equals my number 734,413. 863 itself is prime but 851 is not. Using the same method for 851 we see that 851 is 49 away from 900 (perfect square of 30*30) so its factors are 30+7 and 30-7, so 23 and 37. If one factors the original number starting from 1 and using all primes you will not get a 'hit' until you reach 23. There is no luck in using the perfect squares method - it will work or it will not. Heck just use the smaller example of 851. By the perfect squares method you know right away that 851 is not prime because 30*30 is 900 and 900-851 = 49 which is also a perfect square. You will know the factors in a few seconds.
@@allasar If you know the squares from 1 to 100 like the kid said then for some numbers like 2021 it would be faster to get the factors using the perfect squares method because you would not get a hit the other way until you reached 43. Of course with your quadrillion number the perfect squares method would not work and your first divisor found would be 7 . Then there are other numbers like if you add 2 to your quadrillion number- that number would take a LONG time to factor since the factors of that number are two large prime numbers themselves.
For the number 2009 either prime factorization method OR the perfect squares method can be applied and the prime factors found relatively quickly . For a number like 2021 the perfect squares method would be faster since you would have to test several prime factors using the other method before getting a’hit’ with the prime number of 43. Of course there are numbers where the perfect squares method would not work especially if the number is very large but the same argument can be made for the other method too if the large number does not jive with the known divisibility rules
Im currently a grad student but i remember 4-5 years ago I was searching for good quality content just like yours, but to no avail. Im really glad you're filling up that void!
Definitely agree there. I would probably have continued maths in university if there was this level of content. At the time there was only Khan academy (not to insult them it just wasn't in depth enough back then).
Another nice way to solve this: Notice that 41 is a 4k +1 prime, and can be expressed uniquely as the sum of two integers squared, 4² and 5². Thus √2009 = √(7² × 41) = √(7²4² + 7²5²) We can conclude that √2009 is the length of the hypotenuse of a right triangle with sides 28 = 4×7 and 35 = 5×7. Therefore, all the possible solutions of √a + √b = √2009 correspond to all possible combinations of integer sided* right triangles that fit inside the 28 × 35 triangle. *EDIT: by "integer sided" I mean that every side of the triangles BUT the hypotenuse must be an integer, obviously (because the hypotenuses have to add up to √2009).
@@ciberiada01 √41 is the hypotenuse of a 4, 5, √41 right triangle, while √1476 is the hypotenuse of a 24, 30, √1476 right triangle. Both have slope 35/28=30/24=5/4. Finally 4+24=28 and 5+30=35, so it checks out.
@@valeriobertoncello1809 Oh, what an elegant solution! Thank you, Valerio! 👏👍 I just didn't understand it at first. So, you take the right part: √2009 = √(7²41) 41 is obviously a prime, but because it's 4k + 1 prime, the *only* way to represent it by two perfect squares is: 41 = 4² + 5² {1} And why do you need perfect squares and not just *any* numbers? Because in this way, you can represent √41 as the right triangle's hypothenuse (apply the Pytagorean theorem). The same is valid for √(7²41) : √(7²41) = √(7²(4² + 5²)) = √(28² + 35²) And this is the hypothenuse of our "wrapping" triangle. Its sides are 28 and 35. ❕With {1}, we are sure it exists only one such triangle. Now, you do the same for √a and √b So, √a represents another right triangle's hypothenuse: √a = √(m² + n²) Same goes for √b : √b = √(p² + q²) , where m, n, p, q are the sides of these 2 smaller right triangles. So we have: √(m² + n²) + √(p² + q²) = √(28² + 35²) I imagined that if you align the 2 triangles, so that their hypothenuses √a and √b follow the same line, you get: √a + √b = √2009 (hypothenuses) m + p = 28 = 4×7 (first sides) n + q = 35 = 5×7 (second sides), But not any m and p between 0 and 28 will do! Because all sides must be integer and we must keep the same 5/4 slope, m and p must be multiples of 4, as well as n and q must be multiples of 5. Thus, there are exactly 8 pairs that satisfy this: m, p, n, q, √a 0 28 0 35 0 4 24 5 30 √41 8 20 10 25 √164 12 16 15 20 √369 16 12 20 15 √656 20 8 25 10 √1025 24 4 30 5 √1476 28 0 35 0 √2009
@@ciberiada01 Yes, exactly! I was inspired by 3b1b's video on π/4 = 1 + 1/3 - 1/5 + 1/7 ... that talks about Gaussian Integers and complex factoring of Natural numbers. Really good stuff! Here's the link: th-cam.com/video/NaL_Cb42WyY/w-d-xo.html
It is mathematics, not mathematic. So it's maths not math. Similarly, physics is a singular word but no one tries to drop the s. After all that who cares! The math is wonderful however we spell or say It!
Conclusion that sqrt(2009a) is an integer is wrong. You can only say that it is 1/2*integer or that 2*sqrt(2009a) is an integer Also the same mistake is made with 7sqrt(41a). My opinion that this solution will be enough for 5/7 points
Yea I see what you mean. Just because 2 times the root has to be an integer doesn't mean that the root has to be one, it could be a fraction where the 2 cancels the denominator.
Wait in hind sight I think I understand it now. I think there is a proof that states that a square root is either an integer or an irrational number. Because you can't multiply an irrational number with an integer to get anything other than another irrational number, the root has to be an integer. I wish he would have said that for even 5 seconds, it would have cleared up a lot of misconceptions.
@@randomdude9135 argument inside the square root is not divisible (coprime with) (prime number)^2, i.e. you can have as many distinct prime numbers in the prime factorisation as you want but their powers must all be exactly 1
Useful fact to remember for 4- or 5- digit number factoring: 1001 = 7*11*13 (in problems that are set up nicely with small-ish prime factors). To test 2009, just test 2009 - 2*1001 = 7 to see 7 is a factor. Even for, say, 15877 it's still pretty good: just check 15877 - 15015 = 862 to get 7,11,13 out of the way in one go. Or even better 16016-15877 = 139
Easier test is to test 2009 for divisibility of 2, 3, 5, 7, ... Divisibility is easily tested by adding modulo each digit by "weight". Weights for 2 is 0, 0, 0, 1, for 3 is 1, 1, 1, 1, for 5 is 0, 0, 0, 1, for 7 is -1, 2, 3, 1, 11 is -1, 1, -1, 1, for 13 is -1, -4, -3, 1.
I loved solving this problem on my own! Especially as I'm reading The Art & Craft of Problem Solving, it was really fun to play around with the equation and create various cases (like a=b), figuring out why they don't work, squaring it and getting deeper insight into the conditions the numbers have to meet, and finally going back and looking at my factors and realizing sqrt(2009) could be rewritten as 7*sqrt(41).
You can say “it follows that sqrt(2009a) is an integer” if only you have proof “sqrt(n) not rational, for any integers n when n is not perfect square” Without it u can only say “ 2sqrt(2009a) is an integer”
We have the equation b = 2009 - 2sqrt(2009a) + a. We know that b is an integer so then it follows, because both 2009 and a aren't rationals, that 2sqrt(2009a) is an integer divisible by 2 ----> sqrt(2009a) is an integer.
a and b are integers. So, sqrt(2009a) has to be an integer. What am I missing? If 2sqrt(2009a) is an integer but sqrt(2009a) is not, then "a" cannot be an integer, right?
I'm late to the game, but this video just popped up as a suggestion. Why would sqrt(2009a) be an integer? Couldn't it be some integer + 0.5. I think you mean 2*sqrt(2009a) is an integer. Similarly, 7*sqrt(41a) rather than sqrt(41a).
Why do x and y have to be non negative? I thought only a and b need to be non negative. Based on how you defined a and b a negative x or y will still give you a non negative a or b respectfully. I think the reason should have been realizing that sqrt(a) and sqrt(b) will always be positive based on your definitions of a and b. So that would mean that x,y>7 would cause a,b>2009 which is not the case.
If you take a look at the original problem, the maximum value of a and b are 2009. Since they are in the form of 41x^2, you may notice that x^2 must be less or equal to 49. You probably think, well, x can still be negative. Let's see it: if x = -7, then y = 14 but 14^2 > 49 If x = -1, then y = 8 and 8^2 is also > 49. So that's why x and y must be in [0, 7]
@@djvalentedochp x and y can still be negative, the equation x + y = 7 is written assuming x and y are positive. The correct equation should be |x| + |y| = 7, which allows for negative x and y. So the main point is sqrt(x^2) = |x| and not just x.
Your method is somewhat good but it can be expanced to higher level problems.. for example if the question was:sqrt(a)+sqrt(b)+sqrt(c)+sqrt(d)=7 then we can't do it such a way done by you.. So here is almost same method but with different approach》 Write sqrt(2009) as 7.sqrt(41) As done by u x+y=7 ..we can extend it to my q and can say x+y+z+w=7 Now using pnc no. Of non integral solution are :(7+4-1)C(4-1) i.e 10C3. Similarly the ans of your's q must be (7+2-1)C(2-1) i.e 8C1 i.e 8
If a is not necessary integer then sqrt(2009*a) could be in the form X + 0.5, where X is positive integer, so that 2*sqrt(2009*a) is integer. But a is an integer so sqrt(2009*a) can't be of the form X + 0.5, so sqrt(2009*a) must be integer.
after 2:40 the conclusion is that sqrt(2009a) is an integer, but is the correct conclusion not that 2xsqrt(2009a) is an integer because the term in the equation is 2 times the sqrt. And of course after that it is simpel to prove that the sqrt itself is an integer.
@@hansisbrucker813 2sqrt(2009a) is integer, so it is even or odd. If 2sqrt(2009a) is even then sqrt(2009a) is an integer and we can follow Michael. If 2sqrt(2009a) is odd then we have 2sqrt(2009a)=2p+1 with a and p integer. This gives: 8036a=4p^2+4p+1 with a and p integer. But that means even = odd. So we can continue with sqrt(2009a) is an integer. etc.
@@hansisbrucker813 from what we see at 2.40 it follows that 2sqrt(2009a) = a-b-2009 and because a and b are integers the right hand side is an integer and so is 2sqrt(2009a) From my earlier answer you can then conclude that sqrt(2009a) has to be an integer as well. Because from 2sqrt(2009a) is an integer it followed that it has to be an even integer so sqrt(2009a) is an integer as well.
@@2mk4tom11 because some people are salty that they couldn't think this up themselves XD they don't appreciate the beauty of math rather concern themselves to emotionless problem solving alone
@@manamritsingh969 Yeah exactly, in order to solve these higher level problems, you need to be more creative and think about all these things... it’s so amazing people can come up with this stuff and I’m able to witness it. Math is really something special, I agree.
@@2mk4tom11 thanks man for this flowery words :) Yeah, man. The true geniuses are those who THINK UP those problems. This is often more poetic than solving it. A good logic mind can almost solve this but nowhere near could think up such problems, much less the proof problems.
can't we deduce that, because sqrt(2009) is the sum of 2 sqrts, then 2009 **must** have a square factor to allow the split into 2 sqrt parts. Therefore sqrt(a) + sqrt(b) = sqrt(2009) = p.sqrt(N) + q.sqrt(N) | where 2009 = N*(p+q)`2. So we look for the square factor of 2009 - and bob's your uncle. / finding that square factor could be a bit of a trek tho .. it's anywhere in the squares of odd nos 3..43
Hi, I don't understand how it follows naturally from the equation that sqrt(2009a) should be an integer.. It can also be a decimal like 1.5 so that 2*sqrt(2009a) is an integer
That is because sqrt(2009a) can either be a natural number ( if 2009a is a perfect square) or an irrational number ( if 2009a is not a perfect square) but sqrt(2009a) can never be a terminating decimal like 1.5
@@siralanturing9103 well, this is a negligible problem. Just think of 2009 units. You don’t need to care if that’s feet, nm, miles or mikrometer. Just the relation between the number counts, not what the exact kind of unit Someone might use. The described relationships are invariant with respect to all units…
@@howmathematicianscreatemat9226 Yeah, I know. What I meant was imagine if we had a hypotenuse of 2009 m and we were told to find the sides in cm. That would've been something, no?
I'm Korean Student and I have a similar but easier for coming out my brain. because setting a as 41*x^2 is not thinkable way for most people. a = 2009 + b - 2*sqrt(2009*b) -> a should be integer, so 2009*b must be a square form of an integer k. 2009*b = k^2. 2009 is 7^2*41, in order to satisfy the condition, b should be 0, 41, 41*2^2, 41*3^2 , ... , 41*7^2 which is 2009.
Thank you, but I came the solution simply factoring 2009 which given 7.√(41) and then it is not difficult to figure out that √(a)+√(b) should be of the following format x√(41)+y√(41) with x+y=7and you have the possibilities as described in your video. Thanks again
Generating 41*49 = 2009 is very meaningless. Is there any way for competitors can calculate this by hand in competition? I think the only thing that makes the question complex is just knowing 41*49 = 2009 this equality. If I am wrong please correct me.
3:02 it follow that -2sqrt{2009a} is integer. But yeah from that it follows sqrt{2009a} is integer because √n is either irrational or integer if n is integer is a thm
Or, to do the multiplication and give the actual numbers, four of the solutions are a=0, b=2009; a=41, b=1476; a=164, b=1025; and a=369, b=496. The other four solutions are found by swapping the values of a and b in these four pairs.
Well, he is actually already a genius, maybe not to all mathematicians, but certainly to the average person. He is an Olympiad winner. So this means he doesn’t really work, he is like a boat on the river of creativity. He pushes the door open with his feet and finds short cuts. He doesn’t do unnecessary calculations, in fact, he is the artist of minimizing the amount of calculations in a problem…
Great question! Lets look at the board at 7:23. Notice that even if we used the negative counterparts of x and y we are going to take the square root of (x^2) and (y^2). And sqrt(x^2) = |x| because we use the positive answer by convention (the question itself is phrased using principal roots). Ultimately its just a choice to take positive integers because the important idea is that a,b are 41 times a perfect square.
We have sum of integer plus square root equal integer number. It means that square root is integer number really. For example, 3+5+x=17 obviously can't be truth if x isn't integer number. So '2009a' is a square number (because our square root is integer). 2009=41×49=7²×41. So 41a is a square number and it's true only if a=41x² (x is integer).
After seeing √2009=7*√41 we know the sum √a + √b has to be of the form x*√41 + y*√41, since a and b are integers, we know x and y need to be integers too, and their sum x + y = 7.
@Андрей Босой You can check that your answer does not work by either plugging in some values and using a calculator or by √(n²) + √(2009 - n)² = n + 2009 - n = 2009 != √2009.
Your first advice was to factorize. Doing that we note that 2009 = 49*41 -> sqrt2009) = 7*sqrt(41). Substituting this in the given equation -> sqrt(a)+sqrt(b)=7*sqrt(41). Put a = m^2*41 and b= n*2*41 -> m*sqrt(41) + n*sqrt(41)=7*sqrt(41) -> m + n = 7 A table can now be constructed putting m=1,2,3,4,5, a = m^2*41, b= ( 7 - m )^2*41 ....
Not quite, as you're assuming that m+n being an integer would imply m and n are both integers, which isn't necessarily true, and you haven't provided any proof elsewhere that m and n should be integers, so as it stands, this proof is incomplete, or misleading, even.
@@crustyoldfart I don't follow - where's the stipulation? It certainly isn't obvious from anything you've stated that m and n should be integers. You can't just choose them to be, since then you're finding some, not necessarily all solutions. That's why you need to prove they must be integers, not say they are.
Farhan Awais. OK sport, have it you own way. I'm not about to spend time dealing with your objections. The real takeaway for this piece is that if, as is in the problem as presented, the sum of two radical is and integer, then that integer must be the product of a prime and a square.
Hi Michael, Your intuition about x and y is not correct. There is no constraint about they are beeing positive or non-negative. x squared and y squared are naturally positive. In fact at 6.22 there is nothing to notice. They can be negative and they have to be covered as absolute values ( |x| and |y| ). For the chart at 6.47 chart should be for abs(x) and abs(y) in fact. In brief; there is ∞ possible (x,y) combination. 8 of them holds the criterion that |x|+|y|=7. Best Regards
I solved it in very short and simple way : Sqrt A + Sqrt B = 7sqrt 41 => Sqrt A/41 + sqrt B/41 = 7. Since 7 is integer therefore a = 41k^2 and b=41m^2 => k + m = 7. Hence k,m = (0,7) (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) and (7,0) which brings us to the final answer.
The only condition is that a and b must be positive but x and y can be negative. Since a = 41x^2 and b = 41y^2, therefore |x| + |y| = 7. Am I missing something here?
Whether we take the values of x & y both negative and positive here, the values will be the same for a and b. That's why it was not considered as an issue to be discussed.
Since 'a' and 'b' must be integers, as the exercise wants, 41a will be also an integer. Thus, sqrt(41a) will be an integer or a irrational number. It can't be a rational decimal like you proposed. Please, correct my argument if you find it wrong. Also, feel free to correct my English since it is not my first language and I'm willing to learn.
@@marcosgomes7363 we know that (2 * sqrt(2009 a)) is an integer. Given that information we could say that sqrt(2009*a) could be a multiple of 0.5 but I don’t see how we can say sqrt(2009*a) is an integer itself
@@keco185 Since we're considering from the beginning that 'a' and 'b' are integers, the product 2009*a is also a integer and thus sqrt(2009*a) can be either a integer, if 2009*a is a perfect square, or a irrational number (possibility that is discarded since the product 2 * sqrt(2009 a) must be a integer). So, yes, we can't say that sqrt(2009*a) is an integer only knowing that (2 * sqrt(2009*a)) is an integer. But we can say it knowing that 'a' is also an integer.
b = 2009 -2√(2009 a) + a So, a - b + 2009 = 2 × √(2009 a) So, we know that here L.H.S. is an integer, since a, b have to be an integer (in question:- non-negative integer). Now, dividing both the sides with 2 will give us √(2009 a) = ( a - b + 2009 ) / 2. So, this can be concluded that:- √(2009 a) is only an integer if ( a - b + 2009 ) is an even number.
@@tanmayshukla5330 But only "a" has to be an integer, not √(2009 a). It could be something with an half if ( a - b + 2009 ) is an odd number. The point is, that "a" has to be an integer, so 2009a has to be one too. But the square root of an integer can't end with an .5, because for that the square of a .5 number has to be an integer. A .5 number squared will always end with .25, so it cant be integer.
Another way of seeing this problem is that if you think about it, two square roots when added make one term only if they can be broken down and reduced to a form where the value inside the square root is the same in both terms. (a.k.a multiples of one variable) so you could reduce the given equation to y(c)^1/2+x(c)^1/2=(2009)^1/2 (where x and y are some variable and c is a factor of 2009 whose square root cannot be broken down or simplified) that reduces to (x^2+y^2)(c)=2009 and with the factors, it should be easy. Note I haven't actually double-checked (I am lazy) but I don't see anything wrong with this and if there is something wrong I would love to know about it.
Let N=41a. The square root of N is an integer, which is to say N is a perfect square. Since 41 is prime, it cannot be broken into the product of smaller numbers. For N to be a square, it must come from squaring something which already contains 41, that is, N is the square of [41 times something]. Letting x be that something, from N=41a and N=(41x)^2 we deduce a=41x^2.
One nit I have with a lot of these videos is that they spend so much time laboriously talking through minor things like set notation and being super explicit, but then just kinda hand-wave over more interesting pieces of the proof like (3:47) sqrt(41a) integral => a = 41x^2 This is because any perfect square must have all even powers in its prime factorization, because for any n = p_1^(e_1)...p_k^(e_k), n^2 = p_1^(2e_1)...p_n^(2e_n). So in order for 41a to be a perfect square, a must contain at least one factor of 41 to make 41's exponent even, and then all the rest of a's prime exponents must be even as well (which implies that the rest of a's primes taken alone are themselves a perfect square).
So if you consider the following: sqrt(p_1)+sqrt(p_2)≠sqrt(a) where p_1 and p_2 are distinct prime numbers and a is a nonnegative integer not divisible by p_1 or p_2, then is that statement true? I think it is, but I don’t know if there is some clever or obvious solution that I am overlooking.
Non-trivial solutions exist for n only when n is an integer multiple of a square other than 1. The prime factorisation of 2021 is 43*47, so 2021 has no solutions other than a=0, b=2021 or vice versa.
I just found out that √2009 equals 7√41 and found all the combinations that make up that number(√41 + 6√41, 2√41 + 5√41, etc). And if √a is 2√41, then a is 4*41, and b is 25*41
Since √a + √b = 7*√41, it follows that both √a and √b must be multiples of √41. So we can divide through by √41, giving √(a/41) + √(b/41) = 7. Since both terms now on the LHS must be non-negative integers, they must come from the set { (0,7), (1,6), (2,5), (3,4) } and their reversals. Call them (n, 7-n) for 0
The first thing that came into my mind was a=0 , b=2009 lol
you gotta find *all* positive integers
I too. Same
Arnav Singh
did I say it was, euler?
@@heh2393 it says all non-negative, which includes 0 too
Lol
"That's a good place to stop" is the new Q.E.D
And "okay, great" is the new implies sign(=>).
TAGPTS
LMAO
Quod erat demonstandum
@@danielalp6871 WoW
I don't know if this question is too easy for an Olympiad or if your channel is so awesome that questions like these are becoming second nature.
On second thought, I could totally see this as a highschool math contest problem.
This is definitely one of the easier questions. But a nice one!
@@hassanakhtar7874 This looks like one of the harder (but not hardest) Euclid problems.
It's very easy for a BMO Round 2 problem
it is easy to solve but hard to see. unless you encountered with similar problems before
Ah, yes, 49*41, of course...
2009 ends with 9, that's the hint
@@allasar For much larger numbers the way BeeBee does it will be faster - take the number 734,413. The next perfect square that is bigger than this number is 857*857 = 734,449, Notice that 734,449 - 734,413 = 36 which is also a perfect square and so the factors of my number are 857-6 and 857+6 = 851 and 863. 851 * 863 equals my number 734,413. 863 itself is prime but 851 is not. Using the same method for 851 we see that 851 is 49 away from 900 (perfect square of 30*30) so its factors are 30+7 and 30-7, so 23 and 37. If one factors the original number starting from 1 and using all primes you will not get a 'hit' until you reach 23. There is no luck in using the perfect squares method - it will work or it will not. Heck just use the smaller example of 851. By the perfect squares method you know right away that 851 is not prime because 30*30 is 900 and 900-851 = 49 which is also a perfect square. You will know the factors in a few seconds.
@@allasar If you know the squares from 1 to 100 like the kid said then for some numbers like 2021 it would be faster to get the factors using the perfect squares method because you would not get a hit the other way until you reached 43. Of course with your quadrillion number the perfect squares method would not work and your first divisor found would be 7 . Then there are other numbers like if you add 2 to your quadrillion number- that number would take a LONG time to factor since the factors of that number are two large prime numbers themselves.
notice that 2009 = 2025-16 = 45^2-4^2 so (45+4)(45-4)=(49)(41)
For the number 2009 either prime factorization method OR the perfect squares method can be applied and the prime factors found relatively quickly . For a number like 2021 the perfect squares method would be faster since you would have to test several prime factors using the other method before getting a’hit’ with the prime number of 43. Of course there are numbers where the perfect squares method would not work especially if the number is very large but the same argument can be made for the other method too if the large number does not jive with the known divisibility rules
Im currently a grad student but i remember 4-5 years ago I was searching for good quality content just like yours, but to no avail. Im really glad you're filling up that void!
Definitely agree there. I would probably have continued maths in university if there was this level of content. At the time there was only Khan academy (not to insult them it just wasn't in depth enough back then).
Another nice way to solve this:
Notice that 41 is a 4k +1 prime, and can be expressed uniquely as the sum of two integers squared, 4² and 5².
Thus √2009 = √(7² × 41) = √(7²4² + 7²5²)
We can conclude that √2009 is the length of the hypotenuse of a right triangle with sides 28 = 4×7 and 35 = 5×7. Therefore, all the possible solutions of √a + √b = √2009 correspond to all possible combinations of integer sided* right triangles that fit inside the 28 × 35 triangle.
*EDIT: by "integer sided" I mean that every side of the triangles BUT the hypotenuse must be an integer, obviously (because the hypotenuses have to add up to √2009).
And how does (41, 1476) fit in your 28×35 triangle? 😮
@@ciberiada01 √41 is the hypotenuse of a 4, 5, √41 right triangle, while √1476 is the hypotenuse of a 24, 30, √1476 right triangle. Both have slope 35/28=30/24=5/4. Finally 4+24=28 and 5+30=35, so it checks out.
@@valeriobertoncello1809
Oh, what an elegant solution! Thank you, Valerio! 👏👍 I just didn't understand it at first.
So, you take the right part:
√2009 = √(7²41)
41 is obviously a prime, but because it's 4k + 1 prime, the *only* way to represent it by two perfect squares is:
41 = 4² + 5² {1}
And why do you need perfect squares and not just *any* numbers? Because in this way, you can represent √41 as the right triangle's hypothenuse (apply the Pytagorean theorem).
The same is valid for √(7²41) :
√(7²41) = √(7²(4² + 5²)) =
√(28² + 35²)
And this is the hypothenuse of our "wrapping" triangle. Its sides are 28 and 35.
❕With {1}, we are sure it exists only one such triangle.
Now, you do the same for √a and √b
So, √a represents another right triangle's hypothenuse:
√a = √(m² + n²)
Same goes for √b :
√b = √(p² + q²) ,
where m, n, p, q are the sides of these 2 smaller right triangles. So we have:
√(m² + n²) + √(p² + q²) = √(28² + 35²)
I imagined that if you align the 2 triangles, so that their hypothenuses √a and √b follow the same line, you get:
√a + √b = √2009 (hypothenuses)
m + p = 28 = 4×7 (first sides)
n + q = 35 = 5×7 (second sides),
But not any m and p between 0 and 28 will do! Because all sides must be integer and we must keep the same 5/4 slope, m and p must be multiples of 4, as well as n and q must be multiples of 5. Thus, there are exactly 8 pairs that satisfy this:
m, p, n, q, √a
0 28 0 35 0
4 24 5 30 √41
8 20 10 25 √164
12 16 15 20 √369
16 12 20 15 √656
20 8 25 10 √1025
24 4 30 5 √1476
28 0 35 0 √2009
@@ciberiada01 Yes, exactly! I was inspired by 3b1b's video on π/4 = 1 + 1/3 - 1/5 + 1/7 ... that talks about Gaussian Integers and complex factoring of Natural numbers. Really good stuff!
Here's the link: th-cam.com/video/NaL_Cb42WyY/w-d-xo.html
@@valeriobertoncello1809
👍 Really interesting topic!
British 'Math' Olympiad? No way mate, it's 'maths' here.
math isn't a plural noun dude
Not always, though maths is more common. But I am British and was taught math not maths.
I speak Englishs
It is mathematics, not mathematic. So it's maths not math. Similarly, physics is a singular word but no one tries to drop the s. After all that who cares! The math is wonderful however we spell or say It!
Have to laugh at these Americans telling me, a brit with a maths degree, that I'm wrong about my own language in my own country!
When he says, "That's a good place to stop" by God he means it.
I am so happy that I found this channel. Hopefully you won't stop post the new videos🤗
dividing both the sides by √(41) one gets
√(a/41)+ √(b/41)= 7
the unordered pair
(a,b) € { (0, 41*49 },(41,, 41*36 },
(41*4, 41*25 },(41*9 41*16 }}
TH-cam recommended brought me here. This is some interesting stuff I’ll say.
ah yes the fundamental theorem of arithmetic, my favorite theorem
Conclusion that sqrt(2009a) is an integer is wrong. You can only say that it is 1/2*integer or that 2*sqrt(2009a) is an integer
Also the same mistake is made with 7sqrt(41a).
My opinion that this solution will be enough for 5/7 points
Yea I see what you mean. Just because 2 times the root has to be an integer doesn't mean that the root has to be one, it could be a fraction where the 2 cancels the denominator.
Wait in hind sight I think I understand it now. I think there is a proof that states that a square root is either an integer or an irrational number. Because you can't multiply an irrational number with an integer to get anything other than another irrational number, the root has to be an integer. I wish he would have said that for even 5 seconds, it would have cleared up a lot of misconceptions.
2009a is always an integer as a is an integer. So that means √2009a is either irrational or an integer
Indeed, if sqrt(a)+sqrt(b)=sqrt(c), then a, b, and c have the same squarefree part. (provided these numbers are naturals ofc)
Good point!
Wdym squarefree part?
@@randomdude9135 argument inside the square root is not divisible (coprime with) (prime number)^2, i.e. you can have as many distinct prime numbers in the prime factorisation as you want but their powers must all be exactly 1
Useful fact to remember for 4- or 5- digit number factoring: 1001 = 7*11*13 (in problems that are set up nicely with small-ish prime factors). To test 2009, just test 2009 - 2*1001 = 7 to see 7 is a factor. Even for, say, 15877 it's still pretty good: just check 15877 - 15015 = 862 to get 7,11,13 out of the way in one go. Or even better 16016-15877 = 139
This is exactly what I was thinking!
Easier test is to test 2009 for divisibility of 2, 3, 5, 7, ... Divisibility is easily tested by adding modulo each digit by "weight". Weights for 2 is 0, 0, 0, 1, for 3 is 1, 1, 1, 1, for 5 is 0, 0, 0, 1, for 7 is -1, 2, 3, 1, 11 is -1, 1, -1, 1, for 13 is -1, -4, -3, 1.
you should do BMO2 2017 problem 2, it fits your style of videos very nicely. it was also in an IMO shortlist I can't remember the year though.
"Reduce to smaller problem"
Informatics guy: interesting
ahahah
I loved solving this problem on my own! Especially as I'm reading The Art & Craft of Problem Solving, it was really fun to play around with the equation and create various cases (like a=b), figuring out why they don't work, squaring it and getting deeper insight into the conditions the numbers have to meet, and finally going back and looking at my factors and realizing sqrt(2009) could be rewritten as 7*sqrt(41).
the ad was timed perfectly with the snap
I actually solved one of these problems for myself for a change. So glad I found this channel.
You can say “it follows that sqrt(2009a) is an integer” if only you have proof “sqrt(n) not rational, for any integers n when n is not perfect square” Without it u can only say “ 2sqrt(2009a) is an integer”
I was confused by that too - we know "2*sqrt(2009a)" is an integer, which allows that "sqrt(2009a)" could be a half-fraction, right?
TheBrownMotie yes , but 2009a is integer, and if sqrt(2009a) = m/2 (m integer) => sqrt(n) is rational (n integer, and not square)
We have the equation b = 2009 - 2sqrt(2009a) + a. We know that b is an integer so then it follows, because both 2009 and a aren't rationals, that 2sqrt(2009a) is an integer divisible by 2 ----> sqrt(2009a) is an integer.
a and b are integers. So, sqrt(2009a) has to be an integer. What am I missing? If 2sqrt(2009a) is an integer but sqrt(2009a) is not, then "a" cannot be an integer, right?
@@andreybyl So sqrt(2009a) is rational, but not necessarily an integer? That's where I'm confused
Came here to learn maths, left wanting that Dune shirt.
I can't believe the Mentats wouldn't be sponsoring this content! (Although CHOAM does admittedly have a lot of money.)
Michael always knows when it's a good place to stop.
I'm late to the game, but this video just popped up as a suggestion. Why would sqrt(2009a) be an integer? Couldn't it be some integer + 0.5. I think you mean 2*sqrt(2009a) is an integer. Similarly, 7*sqrt(41a) rather than sqrt(41a).
Yeah I had the same thought.
Always thankful for the hints !
Way more efficient than my solution. I spent two notebook pages to figure out that a = 41*q^2
Your method of solving the problem is interesting !!! I enjoyed a lot !
Why do x and y have to be non negative? I thought only a and b need to be non negative. Based on how you defined a and b a negative x or y will still give you a non negative a or b respectfully. I think the reason should have been realizing that sqrt(a) and sqrt(b) will always be positive based on your definitions of a and b. So that would mean that x,y>7 would cause a,b>2009 which is not the case.
If you take a look at the original problem, the maximum value of a and b are 2009. Since they are in the form of 41x^2, you may notice that x^2 must be less or equal to 49. You probably think, well, x can still be negative. Let's see it: if x = -7, then y = 14 but 14^2 > 49
If x = -1, then y = 8 and 8^2 is also > 49. So that's why x and y must be in [0, 7]
@@djvalentedochp thank you, pls post this as a comment, I was breaking my head trying to figure this out
@@djvalentedochp x and y can still be negative, the equation x + y = 7 is written assuming x and y are positive. The correct equation should be |x| + |y| = 7, which allows for negative x and y. So the main point is sqrt(x^2) = |x| and not just x.
8:24
Thanks bro
I almost missed it. Thanks novelty YT account!
You really commited to that joke
@@Lukasek_Grubasek and he got his reward
I made it to 300 ♪
Your method is somewhat good but it can be expanced to higher level problems.. for example if the question was:sqrt(a)+sqrt(b)+sqrt(c)+sqrt(d)=7 then we can't do it such a way done by you..
So here is almost same method but with different approach》
Write sqrt(2009) as 7.sqrt(41)
As done by u x+y=7 ..we can extend it to my q and can say x+y+z+w=7
Now using pnc no. Of non integral solution are :(7+4-1)C(4-1) i.e 10C3.
Similarly the ans of your's q must be (7+2-1)C(2-1) i.e 8C1 i.e 8
I watched the problem solved by different people but I didnt get it. Your explanation is much more clearer.
If a is not necessary integer then sqrt(2009*a) could be in the form X + 0.5, where X is positive integer, so that 2*sqrt(2009*a) is integer. But a is an integer so sqrt(2009*a) can't be of the form X + 0.5, so sqrt(2009*a) must be integer.
after 2:40 the conclusion is that sqrt(2009a) is an integer, but is the correct conclusion not that 2xsqrt(2009a) is an integer because the term in the equation is 2 times the sqrt. And of course after that it is simpel to prove that the sqrt itself is an integer.
Why is that so?
@@hansisbrucker813 2sqrt(2009a) is integer, so it is even or odd. If 2sqrt(2009a) is even then sqrt(2009a) is an integer and we can follow Michael.
If 2sqrt(2009a) is odd then we have 2sqrt(2009a)=2p+1 with a and p integer. This gives: 8036a=4p^2+4p+1 with a and p integer. But that means even = odd.
So we can continue with sqrt(2009a) is an integer. etc.
@@henkhu100 Why is it obvious it is an integer?
@@hansisbrucker813 from what we see at 2.40 it follows that 2sqrt(2009a) = a-b-2009 and because a and b are integers the right hand side is an integer and so is 2sqrt(2009a)
From my earlier answer you can then conclude that sqrt(2009a) has to be an integer as well. Because from 2sqrt(2009a) is an integer it followed that it has to be an even integer so sqrt(2009a) is an integer as well.
@@henkhu100 Ooooh I totally missed that.
The explanation of this problem is just ingenious.
@jqbtube Why not?
@@2mk4tom11 because some people are salty that they couldn't think this up themselves XD they don't appreciate the beauty of math rather concern themselves to emotionless problem solving alone
@@manamritsingh969 Yeah exactly, in order to solve these higher level problems, you need to be more creative and think about all these things... it’s so amazing people can come up with this stuff and I’m able to witness it. Math is really something special, I agree.
@@2mk4tom11 thanks man for this flowery words :)
Yeah, man. The true geniuses are those who THINK UP those problems. This is often more poetic than solving it. A good logic mind can almost solve this but nowhere near could think up such problems, much less the proof problems.
can't we deduce that, because sqrt(2009) is the sum of 2 sqrts, then 2009 **must** have a square factor to allow the split into 2 sqrt parts. Therefore sqrt(a) + sqrt(b) = sqrt(2009) = p.sqrt(N) + q.sqrt(N) | where 2009 = N*(p+q)`2.
So we look for the square factor of 2009 - and bob's your uncle.
/ finding that square factor could be a bit of a trek tho .. it's anywhere in the squares of odd nos 3..43
This same kind of problem came in PRMO india ( 1st round of 4 national rounds) in recent years
2:52-2:53, KING CRIMSON HAS ERASED 2 SECONDS
DUDE😭😂😂😂😂😂😂😂😂😂😂😂
Would that be a failed solution if we only gave the (a,b) that are not equal to (b,a) also?
I mean, half the solutions are mirrors of each others...
If you are really lazy you can give the (a,b) solutions then use the phrase "up to permutations".
me knowing that the thumbnail question is hard
also me: Thats a good place to stop
Hi, I don't understand how it follows naturally from the equation that sqrt(2009a) should be an integer.. It can also be a decimal like 1.5 so that 2*sqrt(2009a) is an integer
That is because sqrt(2009a) can either be a natural number ( if 2009a is a perfect square) or an irrational number ( if 2009a is not a perfect square) but sqrt(2009a) can never be a terminating decimal like 1.5
Shivansh sanoria, yes that makes sense. In the equation, it has to be a rational number. Thanks!
so the answer is
41(n^2) w/ 0
Thanks for the lecture. Please do more Britsh Math Olympiad round 2 problems, Thanks in advance.
The first solution that I thought of was pythagorean theorem. 🤦♂️
So a hypotenuse of 2009.. metres? That would be a... great value.
@@siralanturing9103 I mean it could be 2009mm, or cm, don’t see what you’re getting at
@@siralanturing9103 well, this is a negligible problem. Just think of 2009 units. You don’t need to care if that’s feet, nm, miles or mikrometer. Just the relation between the number counts, not what the exact kind of unit Someone might use. The described relationships are invariant with respect to all units…
@@howmathematicianscreatemat9226 Yeah, I know. What I meant was imagine if we had a hypotenuse of 2009 m and we were told to find the sides in cm. That would've been something, no?
You have many interesting lectures on solving math problems for students who are good at math. Thanks. You can add lectures on inequalities.
Beautiful, aesthetic math problem. By the way, is there credit for partial solution in Math Olympiads?
Understood and enjoyed this exercics
He explains it in a way that makes it look easy
I'm Korean Student and I have a similar but easier for coming out my brain. because setting a as 41*x^2 is not thinkable way for most people.
a = 2009 + b - 2*sqrt(2009*b) -> a should be integer, so 2009*b must be a square form of an integer k. 2009*b = k^2. 2009 is 7^2*41, in order to satisfy the condition, b should be 0, 41, 41*2^2, 41*3^2 , ... , 41*7^2 which is 2009.
Understood the solution from your clear explanatIon .Thanks.
2:31 why isn’t it “2 times the square root is an integer”? and as such the square root term is a multiple of half
RIght, so \sqrt(x) = n/2, so x=n^2/4, but x=2009a is an integer, so n is even.
yay I managed to solve it, largely due to watching your videos!
Congratulation for your valuable and useful work.
I tried it for around 10 mins and came back here I got near but it ended up in wierd variable 😭😭😭😭 he solved it in a better way
Thank you, but I came the solution simply factoring 2009 which given 7.√(41) and then it is not difficult to figure out that √(a)+√(b) should be of the following format x√(41)+y√(41) with x+y=7and you have the possibilities as described in your video. Thanks again
I solved it in exactly the same way 😁
what about a=b=502.25?
thats an answer too, wouldnt this mean you miss one more answer and thus wrong
Respect from Bangladesh sir
Does the equation imply that square root of 2009*a is a HALF integer? Since it is immediately multiplied by two…
Thank you, I learnt something new today.
Generating 41*49 = 2009 is very meaningless. Is there any way for competitors can calculate this by hand in competition?
I think the only thing that makes the question complex is just knowing 41*49 = 2009 this equality. If I am wrong please correct me.
3:02 it follow that -2sqrt{2009a} is integer. But yeah from that it follows sqrt{2009a} is integer because √n is either irrational or integer if n is integer is a thm
Solution is straightforward. 41 being a prime, solves everything fast. Directly: a=41x^2, and similarly b. You don't need all that jargon.
Or, to do the multiplication and give the actual numbers, four of the solutions are a=0, b=2009; a=41, b=1476; a=164, b=1025; and a=369, b=496. The other four solutions are found by swapping the values of a and b in these four pairs.
You mean a = 369, b = 656
It follows that x is a positive integer since a is.
No actually negative values for x produce the same values for x² as positive values do.
A man who loves to work. Go on like this, tighten your hands, and tell you well, my brother
Then I'm the man who loves to watch how others work. How about you?
Well, he is actually already a genius, maybe not to all mathematicians, but certainly to the average person. He is an Olympiad winner. So this means he doesn’t really work, he is like a boat on the river of creativity. He pushes the door open with his feet and finds short cuts. He doesn’t do unnecessary calculations, in fact, he is the artist of minimizing the amount of calculations in a problem…
brilliant solution
Why does x and y have to be non-negative if you square them?
Same question.
Great question! Lets look at the board at 7:23.
Notice that even if we used the negative counterparts of x and y we are going to take the square root of (x^2) and (y^2). And sqrt(x^2) = |x| because we use the positive answer by convention (the question itself is phrased using principal roots).
Ultimately its just a choice to take positive integers because the important idea is that a,b are 41 times a perfect square.
They don't 'have' to be non-negative, they are defined to be non-negative.
You're not the guy that sings "Walter Reed", but I like these videos too.
I don't understand that fundamental theorem of arithmetic part with a = 41x^2 . How do we get "a ="?
We have sum of integer plus square root equal integer number. It means that square root is integer number really. For example, 3+5+x=17 obviously can't be truth if x isn't integer number. So '2009a' is a square number (because our square root is integer). 2009=41×49=7²×41. So 41a is a square number and it's true only if a=41x² (x is integer).
After seeing √2009=7*√41 we know the sum √a + √b has to be of the form x*√41 + y*√41, since a and b are integers, we know x and y need to be integers too, and their sum x + y = 7.
@Андрей Босой You can check that your answer does not work by either plugging in some values and using a calculator or by √(n²) + √(2009 - n)² = n + 2009 - n = 2009 != √2009.
Very good simple explanation.
Dear professor may l ask you a question is optimization calculus an exception or a part of the infinitésimal calculus?thanks lot
Great explanation sir 👍
Your first advice was to factorize. Doing that we note that 2009 = 49*41 -> sqrt2009) = 7*sqrt(41). Substituting this in the given equation -> sqrt(a)+sqrt(b)=7*sqrt(41).
Put a = m^2*41 and b= n*2*41 -> m*sqrt(41) + n*sqrt(41)=7*sqrt(41) -> m + n = 7
A table can now be constructed putting m=1,2,3,4,5, a = m^2*41, b= ( 7 - m )^2*41 ....
Not quite, as you're assuming that m+n being an integer would imply m and n are both integers, which isn't necessarily true, and you haven't provided any proof elsewhere that m and n should be integers, so as it stands, this proof is incomplete, or misleading, even.
@@fahrenheit2101 By stipulation m and n ARE integers, so it follows that m + n is also an integer.
@@crustyoldfart I don't follow - where's the stipulation? It certainly isn't obvious from anything you've stated that m and n should be integers. You can't just choose them to be, since then you're finding some, not necessarily all solutions. That's why you need to prove they must be integers, not say they are.
Farhan Awais.
OK sport, have it you own way. I'm not about to spend time dealing with your objections. The real takeaway for this piece is that if, as is in the problem as presented, the sum of two radical is and integer, then that integer must be the product of a prime and a square.
That is really clear. Thanks you are reliable
Hi Michael, Your intuition about x and y is not correct. There is no constraint about they are beeing positive or non-negative. x squared and y squared are naturally positive.
In fact at 6.22 there is nothing to notice. They can be negative and they have to be covered as absolute values ( |x| and |y| ). For the chart at 6.47 chart should be for abs(x) and abs(y) in fact.
In brief; there is ∞ possible (x,y) combination. 8 of them holds the criterion that |x|+|y|=7.
Best Regards
I solved it in very short and simple way : Sqrt A + Sqrt B = 7sqrt 41 => Sqrt A/41 + sqrt B/41 = 7. Since 7 is integer therefore a = 41k^2 and b=41m^2 => k + m = 7. Hence k,m = (0,7) (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) and (7,0) which brings us to the final answer.
The only condition is that a and b must be positive but x and y can be negative. Since a = 41x^2 and b = 41y^2, therefore |x| + |y| = 7. Am I missing something here?
Whether we take the values of x & y both negative and positive here, the values will be the same for a and b. That's why it was not considered as an issue to be discussed.
You're correct mate
Shouldn't x and y be Z, not non negetive Z, since their squares are going to be non negetive any way which means a,b=41x^2 will be non negetive?
Very interesting problem and an elegant solution.
@Андрей Босой You are joking right?
Around 4:20 min, I agree that sqrt(41*a) must be an integer, say x. Then x-sqr must be equal to 41*a as opposed to a = 41*x-sqr. What do you think?
I don’t understand why the leading multiplier before the square root can be removed. An integer times a decimal can be an integer. E.g. 7 * 1/7
Since 'a' and 'b' must be integers, as the exercise wants, 41a will be also an integer. Thus, sqrt(41a) will be an integer or a irrational number. It can't be a rational decimal like you proposed.
Please, correct my argument if you find it wrong. Also, feel free to correct my English since it is not my first language and I'm willing to learn.
@@marcosgomes7363 we know that (2 * sqrt(2009 a)) is an integer. Given that information we could say that sqrt(2009*a) could be a multiple of 0.5 but I don’t see how we can say sqrt(2009*a) is an integer itself
@@keco185 Since we're considering from the beginning that 'a' and 'b' are integers, the product 2009*a is also a integer and thus sqrt(2009*a) can be either a integer, if 2009*a is a perfect square, or a irrational number (possibility that is discarded since the product 2 * sqrt(2009 a) must be a integer).
So, yes, we can't say that sqrt(2009*a) is an integer only knowing that (2 * sqrt(2009*a)) is an integer. But we can say it knowing that 'a' is also an integer.
@@marcosgomes7363 oh I get it. Thanks so much!
@@keco185 My pleasure!
"It follows that sqrt(2009 a) is an integer"
Could it also be an integer plus a half? We're multiplying it by 2 on the equation
Ya that's the same problem I noticed in this bro👍🏻👍🏻
b = 2009 -2√(2009 a) + a
So, a - b + 2009 = 2 × √(2009 a)
So, we know that here L.H.S. is an integer, since a, b have to be an integer (in question:- non-negative integer).
Now, dividing both the sides with 2 will give us √(2009 a) = ( a - b + 2009 ) / 2.
So, this can be concluded that:-
√(2009 a) is only an integer if ( a - b + 2009 ) is an even number.
@@tanmayshukla5330 But only "a" has to be an integer, not √(2009 a). It could be something with an half if ( a - b + 2009 ) is an odd number.
The point is, that "a" has to be an integer, so 2009a has to be one too. But the square root of an integer can't end with an .5, because for that the square of a .5 number has to be an integer. A .5 number squared will always end with .25, so it cant be integer.
My jaw hit the ground at 5:30
That was so clever
5:30?
sir please tell why can't we take a=2009*x*x,b=2009*y*y why prime factorization is required?
I remember this from a Simpsons episode, where Homer wears glasses in a men's washroom.
Lol
The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side.
You are brilliant.
Nice solution. Thanks. It is very useful
Another way of seeing this problem is that if you think about it, two square roots when added make one term only if they can be broken down and reduced to a form where the value inside the square root is the same in both terms. (a.k.a multiples of one variable) so you could reduce the given equation to
y(c)^1/2+x(c)^1/2=(2009)^1/2 (where x and y are some variable and c is a factor of 2009 whose square root cannot be broken down or simplified)
that reduces to
(x^2+y^2)(c)=2009 and with the factors, it should be easy.
Note I haven't actually double-checked (I am lazy) but I don't see anything wrong with this and if there is something wrong I would love to know about it.
That's exactly what the video does.
How did you get a=41x^2? That part confuses me
Let N=41a. The square root of N is an integer, which is to say N is a perfect square. Since 41 is prime, it cannot be broken into the product of smaller numbers. For N to be a square, it must come from squaring something which already contains 41, that is, N is the square of [41 times something]. Letting x be that something, from N=41a and N=(41x)^2 we deduce a=41x^2.
One nit I have with a lot of these videos is that they spend so much time laboriously talking through minor things like set notation and being super explicit, but then just kinda hand-wave over more interesting pieces of the proof like (3:47) sqrt(41a) integral => a = 41x^2
This is because any perfect square must have all even powers in its prime factorization, because for any n = p_1^(e_1)...p_k^(e_k), n^2 = p_1^(2e_1)...p_n^(2e_n). So in order for 41a to be a perfect square, a must contain at least one factor of 41 to make 41's exponent even, and then all the rest of a's prime exponents must be even as well (which implies that the rest of a's primes taken alone are themselves a perfect square).
So if you consider the following: sqrt(p_1)+sqrt(p_2)≠sqrt(a) where p_1 and p_2 are distinct prime numbers and a is a nonnegative integer not divisible by p_1 or p_2, then is that statement true? I think it is, but I don’t know if there is some clever or obvious solution that I am overlooking.
Why is the sqrt in fact a integer? Can someone explain?
i dont know math language but all i know is you are extremely smart
Now don’t chat, its a good time for you to practice the same question for sqrt(a) + sqrt(b) = sqrt(2021).
Why should I?
If you people progress this way , then one day will come , when where you guys will be eating Humans!!
Then it's a good time for you too. So, give me the answer, I'll check if it's correct.
Non-trivial solutions exist for n only when n is an integer multiple of a square other than 1. The prime factorisation of 2021 is 43*47, so 2021 has no solutions other than a=0, b=2021 or vice versa.
Bao, you're late!
Can you explain the step at 4:05? I do not understand why sqrt(41a) being an integer implies that a is of the form 41x^2.
And how do you know that there needs to be a perfect square in there?
I just found out that √2009 equals 7√41 and found all the combinations that make up that number(√41 + 6√41, 2√41 + 5√41, etc). And if √a is 2√41, then a is 4*41, and b is 25*41
Since √a + √b = 7*√41, it follows that both √a and √b must be multiples of √41. So we can divide through by √41, giving √(a/41) + √(b/41) = 7.
Since both terms now on the LHS must be non-negative integers, they must come from the set { (0,7), (1,6), (2,5), (3,4) } and their reversals. Call them (n, 7-n) for 0
So, 2*sqrt(a) being integer does not mean that sqrt(a) is one. sqrt(a) could be some odd number over two. Same goes for the 7 later on, obviously.
But a is an integer, so it follows in both cases.
No. It follows that 2*sqrt(2009a) is an integer. All sqrt(2009a) need be is a multiple of 0.5
that cannot happen, since the root of an integer is either an integer or irrational
Nice qsn
wow love your way to solve qsn michel😊😊☺️