an exponential Diophantine equation.
ฝัง
- เผยแพร่เมื่อ 16 มิ.ย. 2020
- Using the notion of the order of an integer mod n, we find all solutions to a certain exponential Diophantine equation.
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I, for one, salute the return of "This is a good place to stop"
Y =1 x= 2
Yeeah !
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Solve problems spontaneously it will be interesting to see the thought process of a professional mathematician!
Great video as always.
While he could do that, this would create a split between "lucky videos" where he basically gets a nice solution quite fast, and "unlucky videos" where he reaches several dead ends on his attempts, resulting in longer, not so interesting videos. Not to mention the cases where he could get a wrong answer (would surely happen to me).
Ok.
What would you think of releasing two videos, one with the answer as it is now and another one where we see his reflection, where he starts etc.
@@RianKoja He can edit the uninteresting parts so that its more engaging, and it is interesting to see what a mathematicians first approach, even when it is wrong, is.
Liked this a lot. I didn't see where it was heading, and at 11.54 I said to the screen 'wait a minute, when does this stop?'.... and right on cue, Michael said 'this might seem like it's endless, but we're nearly there'...... brilliant
"Now what you wanna notice, is that 3 to the 18 minus 1 is a multiple of 37" - Penn, Michael
Well I want to notice it but I am sure I won't :D
It makes sense though using some principles from number theory
Euler's Theorem shows us that a^phi(n) = 1 mod (n), where phi is the Euler totient function. This implies that all exponents are modulo phi(n), or in other words ord_n(a) is going to be a factor of phi(n)
Since 37 is prime, that means phi(37) = 37 - 1 = 36, thus 3^36=1 (mod 37). We can now take the square root on both sides to get that 3^18 = 1 (mod 37), meaning 3^18-1 = 0 (mod 37)
@@deidara_8598 actually you can use the Fermat's little theorem and obtain straight away that 3^36=1(mod 37) bcs 37 is prime gcd(3,37)=1
@@deidara_8598 Not quite. You get that 3^18 ≡ ± 1 (mod 37), so you have to rule out the possibility that 3^18 ≡ -1 (mod 37). Easiest way that I can think of where you don't need to calculate 3^18 (mod 37) would be to try to show that 3 is a square mod 37. So we check: 3 is not a square, 40 isn't, 77 isn't, 114 isn't, 151 isn't, 188 isn't, 225 is. So in fact 3^{18} ≡ 225^{18} ≡ (15^2)^{18} ≡ 15^{36} ≡ 1 (mod 37). The other way to prove that 3 is a square mod 37 would be to use Quadratic Reciprocity. Since 37 is 1 mod 4, we know that 3 is a square mod 37 if and only if 37 is a square mod 3, which is obviously is.
I thought this was never going to end, and when it ended there were no solutions 😂😂😂😂
I really appreciate the way how you explain complex Mathematics, and that's splendid.
Hi Michael Penn, I really love your channel! I'm a middle schooler in South Korea who loves mathematics. We're not going to school because of coronavirus, so I spend time on youtube watching videos, and the youtube algorithm recommended you! I watched the integer partition series and became interested in your channel so I'm watching all your videos. I really love your explanations, it's very comfortable to understand!
I wish your channnel gets more bigger!!!
+I wonder if my english is correct;;; 😓😓
It is kind of okay @Taxicab 1729
Only instead of more bigger it will be bigger😊😊😂😂😄😄😉
excellent username btw
@@keepitplainsimple1466 Oh Thanks! I made some mistakes on Grammar 😂😂
@@Alphabet576 Thanks 😄😄
@@_er5te Actually I am also a middle school student just in India😄
So I guess we have something common😉
In an actual exam or test environment is it reasonable to solve this? For example knowing that (3^18)-1 is a multiple of 37, or knowing the order base 37 of 7 is 9. Are these derivable on the fly? Or is this just more of a fun problem to work through
Yeah, it's definitely "easy to see" that 3^18-1 is 37k for some k=10,470,824 😂
@Austin yeah this is a viable olympiad problem
Impossible.
Yeah, it’s definitely doable in under 2 minutes. The key is to keep reducing as you go along. So 7^2=49, but 49≡12. Then to calculate 7^3, we only have to multiply 12 by 7 and reduce, since 7^3=7^2 * 7 (so 7^3=12*7=84 ≡ 10). There are other tricks, like when Michael wrote that 3^3≡6 (7). Since 6≡-1 (7), this means that (3^3)^2≡(-1)^2≡1, and thus the order is 6, without even having to compute 3^4 or 3^5 mod 7!
djsmeguk that’s the beauty of working in modular arithmetic, you never have to come close to calculating these sorts of values of k. Checking that one only took me around 30 seconds!
great job, never stop
except when “this is a good place to stop”
Great video. I'm really enjoying all the olympiad/putnam problems.
For those who wanted a bit more motivation for where the numbers come from this is for you. Since we know there is a solution to the equation we want to mod the equation out by a number (a prime power in this case) which will make one of the powers disappear so we can split the problem into checking a finite number of cases and checking a tail of solutions. We want to make sure the tail is further out so we pick up all the solutions in our split so we would want to pick higher powers of 7^2 or 3^3. It turns out that 27 works fine.
Modding the original equation out by 27 gives three possibilities (x,y)=(2,1), or (x,y)=(1,0) or the general case y = 4 mod 9 with x>=3. Next you would want to construct a contradiction with the 7^y power so it's easier to hack it by choosing a modulus N such that 7^9 = 1 mod N just so you can reduce on the number of variables you're juggling. Splitting into primes gives 7^9-1 = 2*27*19*37*1063. We don't get a contradiction from 19 but 37 works.
Using 7^9 = 1 mod 37 and y = 4 mod 9 reduces the original equation to 3^x = -2 mod 37. Cycling through the powers of 3 (you only have to check nine powers before it repeats) shows that there isn't a solution in this scenario. So QED there are only 2 solutions.
Congratulations also on your channel blowing up over recent months :)
Your solution is awesome
Well, particularly Euler's Theorem and his Totient or phi function...
Very easy to understand, but how could I imagine this way ? Great video ! Thank you.
I actually started working on it in a similar way as Prof. Penn does but decided to give up because I did not know when or IF it was going to end 😊
Can you make video on problem solving strategies (how do you attack problems)
Nice! Cool trick of using the ord over and over again.
This was very understandable, and I appreciate that. It worked because there were no solutions > 2, terminating with a contradiction. Suppose I had a similar problem, but with solution(s) to be found. Would a strategy like this reveal them in a reasonable number of iterations? How would one know when it had worked?
Looking for an answer to this too
One of my fav math channel
thank you for making these videos, they're great
Do appreciate his content by liking and sharing as well, he brings amazing math for everyone ❤️.
Could you make a video on another Diophantine equation, namely |x^2-y^3| = 2 for x,y in N, so in other words 26 is the only natural number that lies directly between a perfect square and a perfect cube?
But how did you prove that 3^18 -1 is a multiple of 37???
Fermat's little theorem. Notice 3^36=1 mod 37
Rest check the factors of phi(37)=36.
@@TechToppers thanks
@@abhijeetkumar1028 Mention not.
Thanks man. You are doing a great work here
What subject book does one read to get the tools/techniques used here? e.g. Is any book on Number Theory sufficient? I only know mod by its definition, I've never used it for anything. I followed this video more or less, but it'd never occur to me to use mods to solve it. And this was the first time I'd ever heard of "order" as used here.
I do. I'm newbie and learned congruency around 3-4 months ago. I still have learn quadratic residues and Chinese Remainder Theorem... But I can sense modular arithmetic now due to Mr Michael.
Watch his INMO video. (Indian National Mathematical Olympiad) That's where I got motivation.
Just checked those facts first before trying to solve this. Congruency seems to be very interestingly used for the proof as I proceed (without watching the video until the end yet) haha
What if (e.g.) x=3? Is this covered by the 2nd case?
Sort of. If x=3, y
@@ostdog9385 tnx!
For every x belonging to R, there is at most one y belonging to R so that 7^y+2 = 3^x. In fact, y = log_7 (3^x-2). In particular, for every x > log_3 (2) there is exactly one real y that solves the equation, for every x < log_3 (2) there is no real y that solves the equation. We already found the couples (x,y) = (1,0) and (x,y) = (2,1), so there are no other solutions of the form (1,y) and (2,y).
Well, I could see lot of similar equation, but what about the a^n + b^n = 2^x one? I know, it is not the place to proof, but is it a known "part" of the Fermat equation? Is there any proof?
Great work....i am enjoying all the video...keep it up👍👍👍
Nice problem! Thanks for sharing, sir. What if 2 is replaced by a variable, wud it be easily solved as well? Thank u
Great and fantastic. As always...
At 14:00 couldn't you immediately conclude that 27 cannot divide the RHS (since then 3 should divide a power of three minus 1)?
You always used a factor that is quite small, such as 3^18-1 = 37m, even though quite obviously there can be other factors. Is there any reason for this choice, or do we do this until a contradiction arises?
I’m curious if this could have been approached through continued fractions: Note that, in the set up, an implication would be that 7^y is in some sense “close” to 3^x Looking for approximations to Log(7)/Log(3) would give [1; 1, 3, 2, 1...]. The first few convergent are (1/1, 2/1, 7,4, 16/9, 23/13,...). Taking each numerator x and each denominator y, consider the absolute value of the difference between 3^x and 7^y. Because these are the closest that powers of 3 and 7 that may be found, it would be reasonable to expect solutions only among these. First we have the pair (3,7) with a difference of 4, then (9,7) and a difference of 2 (the second solution here), then (2187, 2401) and a difference of 214. From here, the differences grow very large very fast. Now let 3 and 7 become parameters A and B with A
In the way that it applies continued fractions to logarithms, this is maybe a little off the beaten path of continued fraction theory generally (but see the work of Shanks on this). Looking for values when A^x is close to B^y focuses on x/y as a good approximation of Log(B)/Log(A), namely, as a convergent. to the continued fraction of Log(B)/Log(A). (Here, the first two convergents, 1/1 and 2/1, give, 3^1-7^1 and 3^2-7^1 and the first two differences, 4 and 2.) Yes, Mikhail, the example could be expanded through the use of semi-convergents (intermediate convergents). The differences still appear to increase monotonically. If we take the differences (absolute value) between A^x and B^y where x/y is a convergent,, we get differences (4, 2, 214, 2693114), etc. Filling these in with the results from semi-convergents gives (4, 2, 22, 100, 214, 2876, 2693114), where 22, 100, and 2876 are supplied by the semi-convergents. (And, as you showed, the difference of 22 comes between 27 and 49, i.e., 3^3 and 7^2, where the ratio 3/2 given by the exponents is the first semi-convergent.) I'm wondering whether either sequence, the one of differences arising from convergents only, or of the longer one that supplements the semi-convergents, can sustain no more than one violation of being monotonically increasing (here implying ≤). I'm only guessing that the answer is yes and, furthermore, the although semi-convergents certainly slow the growth of the sequences down, they do not change the result. It may be worthwhile to begin by separating the sequence of "positive" approximations where Ax > B^y from the "negative ones, with the sign reversed and show that both of these are monotonically increasing, then concatenate them back together.
Adam Romanov, I get them the lazy way by typing “Continued Fraction of Log[3,7] into Wolfram Alpha. C.D. Olds, in his wonderful brief introduction into continued fractions, describes an algorithm derived by Daniel Shanks. It’s a modified version of the usual continued fractions algorithm. For 3 and 7, begin by determining a and b such that 3^(a1) < 7 < 3^(a1+1). Aside from the Olds text, there are versions of the algorithm available online, www.exstrom.com/blog/abrazolica/posts/logarithm.html, and, in a revised form, www.numbertheory.org/pdfs/log.pdf. I haven’t located Shanks’s original paper, but I see that Olds is all online, where the algorithm can be found starting on p. 84. www.ms.uky.edu/~sohum/ma330/files/Continued%20Fractions.pdf
Thank you! Please, help to solve 3 ^ x + 7 ^ y = 104 ^ z where x, y, z are natural numbers.
probably mentioned before: 11:26 should read x - 2 instead of x - 12.
Also, I think the cases x= 3 and x>=3, y
For each x there will be at most one y that solves the equation, likewise for each y there will be at most one x. Since all possibilities where x or y is
Thanks for the great video!
At 6:19, you can directly find this by taking mod 7 both sides, the RHS must have a residue of 2 and this happens when the exponent is 2 + 6k for some integer k
Great video!!
Unless I missed something, there is a bit of a shorter way to solve this.
x - 12 = 18m
x = 18m + 12 = 6(3m + 2) which is a multiple of 6
x - 2 = 6k
x = 6k + 2 which is not a multiple of 6, therefore x does not exist
For the primitive roots, follow the Fermat's Little Theorem and just check a few multiples
what about the case where x2 and x>2, y
It is clear that x > y so one of those needs no checking, and further y cant be too much larger than 2x if at all, so you can brute force it, but yeah he should have mentioned it
I had to pause the video to check that 728 is indeed a multiple of 13. How do you know which factor of p^n-1 to use? Is there a general proof that p^x+n=q^y, for p and q fixed primes and n a fixed integer, has only a finite number of solutions? Is there a general method for finding all solutions?
Fermat's little theorem
Brilliant but I have the issue that these multiples like 13 are plucked out of the air. There must have been a lot of trial and error which went into them. Or if not, it would be good to understand how they were selected.
They are prime numbers.
@@millwrightrick1 Well, sure, but what's the reason we have to take 37 after 11 other primes as it's not kinda easy to get to him??
This is how mathematicians do proofs. You do a lot of work, then write it up so that it seems like magic, including setting up the orders at the beginning so you can refer to them later. Lovely! :D
@@TheNextFool
If I become one... I won't do that.
It's disturbing.
Remember Euler's totient function or Fermat's Little theorem?? That's the motivation!
The case 1 should be "x
Beautiful! Btw, you have a typo at 11:26 (saying 2 but writing 12)
The (1,0) and (2,1) are instant inspection solutions.
The hard/tedious part is showing they are the only solutions.
Well done
is there any reason to write: 3^6 = 1 (mod 7) instead of: (3^6) mod 7 = 1 ?
Mathematicians think about it as two integers being "equivalent mod n", and this dividing the integers into n "equivalence classes", where it doesn't matter which member of the equivalence class you use for your operations as far as the equivalence class of the answer you get. Then they prove properties of these equivalence classes. They're actually saying something more like: {..., -4, 3, 10, 17, ...}^6={..., -6, 1, 8, 15, ...}. The "(mod 7)" to the side is telling you what equivalence classes 3 and 1 are shorthand for, and apply to the statement as a whole. (Note that the 6 is an integer, not an equivalence class, because it matters whether you multiply 6 terms or 13.)
The "(3^6) mod 7" notation is doing integer operations that include a "remainder of division" operation, where you evaluate it as 729 mod 7.
A computer science way of thinking of the math form is that you're casting your integers to a limited-range integer type with precise overflow and underflow rules and then using their arithmetic operations.
Great problem. At 11:30, shouldn't that be x - 2 = 18m?
Indeed. Although he says "x-2" while he's writing "x-12" on the board and when he substitutes, he treats it as if it were "x-2", so no harm done. Except it took me a minute to work out where the extra 3^10 went when he did the subsequent substitution.
Can you please tell me how you know 7¹²-1 is a multiple of 19 without a calculator?
Can you prove that Kepler's equation y = x-a*sin(x) where 0
So you mean x-asinx = 0.
Let f(x) = x-asinx.
For any value of a, f(0)=0 so we just need to see if there are any more solutions.
f'(x) = 1-acosx.
Because cosx 0:
1-a
Mind blowing
I can totally get every basic and complex integral, every single equation, I understand concepts of geometry but I still really really struggle with modular mathematics... I don't get a single thing about it. Any good books? Every book in my native language explains that "congruency" thing really differently and I really can't stand how I am the only one not getting it :(
Thx for the suggestions
@ hey, thanks for the response. My native lang is Czech and the Wikipedia link you sent was the only okay source. Still far away from understanding but not that terrible.
I guess I’ll try a bit even if that’s prob a bit optimistic. Two things are congruent mod x if they have the same remainder when divided by x. Also if a is congruent to y (mod x), we can say a=nx+y for some integer n. Everything else is derived from that and I’d say watching videos like this (and I’d say mathologer has a lot of simpler problems with modular stuff) and then making sure u can prove whatever they do using those facts is my recommendation for how to go about it. Hopefully that sorta helped, I can’t say I’m an expert myself but that’s the basics at least. Also an added note all of these numbers have to be integers or I don’t think it’s applicable (I could be wrong tho)
I guess here’s an example of why if a=y (mod x), then a*b=y*b (mod x) (and note I’m using equals signs instead of congruence signs bc that’s a lot of extra effort honestly).
If a=y (mod x), then a=nx+y for some integer n. Then a*b=(nx+y)*b, which, if we distribute becomes a*b=(n*b)x+y*b. Then, since n*b is an integer multiplying x, y*b is the remainder of a*b when divided by x and so we can rewrite this as a*b=y*b (mod x). QED. hopefully that helped
Why does the solution show there are no more solutions for x,y>2 but not for x,y
Give this man a good soup today. He’s a great and creative mathematician.
since x>-3 we could work with powers of 7 mod 27 right from the start I guess
Why didn't you use 9 in the rhs to lhs
Start using 9 in rhs and you will end up with same solution 0 mod 27 congruent to -9 mod 27
Amazing
Thanks for this very nice problem.
I solved it a bit differently though painfully. I hope I didn't make mistakes 🤔:
Once I reached the level (same as this video) where the right hand side is a multiple of 13:
- I factored the left hand side = 7(7^3-1)(A) = 7*9*2*19*A where A = sum(7^3n) with n from 0 to whatever.
- Then I simply calculated A=sum(7^2n) mod (13).
- From n=0 to n=9, I get the following residues: 1,8,9,3,4,11,12,6,7,1
- So at n=9, I get 1 mod(13) and no 0 before.
- This means any new 7^3n I would add to this cummlative sum would simply cycle again.
- A cannot be multiple of 13.
- So no solution for x >2, y>1
I have tried but i didn't get the same residues as given by you at n=3 sum(7^3n) gives zero mod by 13. It is (7^9+7^6+7^3+1) mod 13 is zero.
@@bhanusri3732 Thanks for pointing this out. You are right, my calculation was totally flawed.
sum(7^3n) is divisible by 13 for n=1mod(4).
I re-did the math and I ended up going thru the same stages as in this video up to 7^9n-1 being divisible by 3^3=27 but the other side of the equation only got 3^2 and another term 3^z-1 which cannot divide 3.
So essentially, my initial approach was wrong and an incorrect shortcut. 😞
C est hypnotisant mais il manque la roulade 🤸♂️
Thanks for this great lovely application of congruences. There is a typo at 11:26 : it should be " x - 2 " instead of " x - 12 ". And at 13:23 probably you meant " 9 times n " instead of " 9 to the n ".
You missed the cases where x = 3 and vice versa. They might be covered by the other cases somehow but they should be mentioned at least.
How do you notice that 7^12-1 is a multiple of 19, without a calculator? (At 10:11)
You can factor 7^12-1 as (7^6-1)(7^6+1) = (7^3-1)(7^3+1)(7^6+1) = (7^3-1)(7^3+1)(7^6+1) = (7-1)(7^2+7+1)(7^3+1)(7^6+1). Then you check them one by one and you notice that 7^2+7+1=57=3*19, so you can check the other side of equation mod 19. And if you want to show that 3^18-1 is a multiple of 37 here is how to do it:
3^18-1 = (3^9-1)(3^9+1) = (3^3-1)(3^3+1)(3+1)(3^8-3^7+3^6-3^5+3^4-3^3+3^2-3+1) = (3-1)(3^2+3+1)(3^3+1)(3+1)[3^6(3^2-3+1)-3^3(3^2-3+1)+(3^2-3+1)] =
(3-1)(3^2+3+1)(3^3+1)(3+1)(3^2-3+1)(3^6-3^3+1).You check them and on the last one you see that 3^6-3^3+1 = 27^2-27+1=27*26-1=703=37*19.
What I don't know however is how to decide which number should we choose as next mod number.
Uhh, Fermat's little theorem and some theorems regarding order of a modulo n.
if you could have an equation like this that would go on to infinity (if thats even possible) could you use the fact that infinite primes cant divide a number to get a contradiction?
Bro wtf I already watched this video two years ago?
After 14:25 isn't it enough to say that since the left hand side is a multiple of 27, the right hand side must be too - so 3^(x-2)-1 must be a multiple of three, but it is not - it is one less than a multiple of three.
You could just see that 27=3^3 while 9 is 3^2 and 3^(x-2)-1 is not divisible by 3 for the last part instead of invoking the x-2=6k fact and its shorter and cleaner. Still, cool problem and nice proof.
Can anyone tell me the solution of this functional equation or its source please :)
The problem:
Let α,β ∈ Q* and α≠β .
Find all functions f: R ->R
Such that:
f([x+y]/α)=[f(x)+f(y)]/β
Well I found y = 1 and x = 2 in short order lol
x and y cannot be equal since x, y = 0 fails and y ≥ ln(2)/ln(7) (also fails) since 7^(ln(2)/ln(7)) = 2.
7^y grows much faster than 3^x so x and y must not be far apart if 7^y - 3^x = -2.
Putting everything together, y < x.
Substituting y = 0 we get 7^0 - 3^x = -2, then x = 1
Rearranging, 7^y + 2 = 3^x we see that if y = 1 we get 7 + 2 = 3^2, so x = 2
I finished here because I was unable to prove that these were the only two solutions (And I don't know the first thing about orders)
Solutions: x_1 = 1, x_2 = 2; y_1 = 0, y_2 = 1
Hmm I don't quite see why it wasn't faster: once you get x-12=18m, you get that x is divisible by 3, but just before you had x-2=6k which shows that x=2 mod 3 which is a contradiction.
I don't understand why case 2 excludes x,y
Is it because the the exponents x-2 and y-1 need to be greater than or equal to 1 to be an order m mod n?
There is a case x>2 y
It’s probably easily checkable since one side will probably get way to big rly fast but good point he prob should’ve covered it
First draw graph for 7^y + 2
then draw graph for 3^x
the point of intersection is the solution.
@Adam Romanov : Of course you can do it in 3 dimensions! The solution is x = 27, y = 15.24359... (Not integer though)
Solution by insight
7^y+2=3^x
7+2=9
y=1, x=2
I think its not accurate to say the second case only checks x and y >=3.
It seems to miss the case where one of them is less than 3 and one is greater than or equal to 3. Of course it doesnt. The second case is more like just all possibilities that are not the first case.
"What I want you to notice is that 3^18 - 1 is a multiple of 37."
I mean, yeah, obviously
Fermat's little theorem will give you the intuition.
YOU ARE A GEEZER AND A HALF!!! I wish I could give you more subscriptions... "...this a good place to stAp..."
Is 0 a natural number?
@@angelmendez-rivera351 no its not
y,x = 0,1 and 1,2 were instantly guessed. Proving that there are no other solutions is much harder esp. if you see 2 solutions so quickly without doing any real math !!
Wasn't it enough to prove there were no answers for x,y>2 when we got that y-1 had to be multiple of 9 (always odd) and 12 (always even) at the same time?
It could still be a multiple of 9*21 = 189
Multiple of 9 may be even for example 18 = 2*9 .
36 is a multiple of both 9 and 12
@@36sufchan Agh that's true I haven't thought about that idk why!! Thank you all =)
I think you were thinking of >powers< of 9 being always odd?
Solve 2^a-3^b=7 for all integer numbers a,b, please.
I guessed one solution (2,1)
That's the only solution mate... As (0,1) is not valid
Is it CON-gruent or con-GRUENT?
Both sound OK. (American here) But it really is as if neither is stressed. If one version had to be picked, I'd go with con-Gru-ent
Isn't there a shorter proof for this solution?
@ Yeah there is. Do u wanna know how ?
@@Subhajit_Paul123 sure
Can you see my comment ?
@@Subhajit_Paul123 yes I see your comment
13:22 It is not "nine to the n"
but "nine times n".
Please start lectures on discrete mathematics!
This stuff is sort of easy.
At 11:27, x-2 = 18m not x-12
Bruh, playing around with this function in desmos will boggle your mind...
V. Nice
There is another way to solve this, inspired by "youtube . com/watch?v=RjxDP6jCjZc"
First notice *(x; y) ∈ { (1; 0), (2; 1) }* are the only solutions to *y < 2.*
Consider *y ≥ 2.* This leads to *x ≥ 3,* so the RHS (and thus the LHS) is divisible by *27* and *9.* Use the _Binomial Theorem_ on the LHS. All but the first two terms vanish:
*0 ≡ 7^n + 2 = (1 + 6)^n + 2 ≡ 3 + 6n mod 9 (1)*
*= 3(1 + 2n) mod 9 => n = 1 + 3n', n' ∈ ℕ_0*
Reduce (1) *"mod 27",* using the _Binomial Theorem_ again. Insert *n = 1 + 3n'* and luckily the quadratic term vanishes:
*0 ≡ 7^n + 2 ≡ 3 + 6n + 18n(n - 1) mod 27*
*≡ 9(1 + 2n') + 0 mod 27 => n' = 1 + 3k, k ∈ ℕ_0*
We combine both results and get *n = 4 + 9k, k ∈ ℕ_0.* Reduce both sides of the original equation *"mod 37"* and notice two repeating patterns within the exponentials:
*n | 0 1 2 3 4 5 6 7 8 9*
*3^n mod 37 | 1 3 9 -10 7 -16 -11 4 12 -1 => ord_37(3) = 18*
*7^n mod 37 | 1 7 12 10 -4 9 -11 -3 16 1 => ord_37(7) = 9*
This helps, since *7^9 ≡ 1 mod 37* simplifies the LHS after inserting *n = 4 + 9k:*
*7^n + 2 = 7^4 * 7^{9k} + 2 ≡ (-4) * 1^k + 2 = -2 mod 37*
Another look at the table shows that *3^n* never reduces to *"∓2 mod 37".* The LHS and RHS cannot be equal *"mod 37"* -- but then they cannot be equal to begin with: The only solutions to the problems are the ones from the beginning with *y < 2* !
how can i know 7^12-1 is a multiple of 19 without computer
use math and brain
jk I didn't have a clue either
(7^12-1)=(7^6+1)(7^6-1)
=(7^6+1)(7^3+1)(7^3-1)
=(7^6+1)(7^3+1)(7-1)(7^2+7+1)
=(7^6+1)(7^3+1)*6*57
=(7^6+1)(7^3+1)*6*3*19
Let's begin.
7^3=343=-1(mod 19){Typical Observation}
7^6=1 (mod 19)
7^12=1 (mod 19)
By definition of congruent modulo n, 7^12-1 must be multiple of 19.☺️
Fairly standard.
@@TechToppers tq
Guess and check is a valid problem solving method, so Y = 1 X = 2. Are those the only solutions, don't know but I solved it. Okay let me start the video
Amazingly hard problem
Yikes! It would be nice if you explained how the particular factors at each stage were selected. They seem to have been selected arbitrarily, but the particular choices of course led to a fortuitous convergence (if that's the right word).
Actually, there's a much simpler path to a proof. Considering the equation 7(7^(y-1) - 1) = 9(3^(x-2) - 1) modulo 7, we can conclude that x-2 = 6k (since ord_7(3) = 6). From this, we can conclude that 3^6 - 1 must be a factor of the RHS. But 3^6 - 1 has a factor of 7^2. On the other hand, the LHS cannot have a factor of 7^2 if y > 1, since the quantity 7^(k-1) - 1 and 7 are relatively prime. That proves that there are no solutions with y > 1 and we are done.
But why this method will not work for smaller x and y
If x and y are negative then 7^y and 3^x would be fractions that can't have a common denominator.
@@toriknorth3324 for smaller x and y means x, y€{0, 1,2}
Why you checked 0 solution? X and Y are is natural, but not an integer!
Y =1 x=2
x=2 y=1
How did I guess that there were no other solutions!!!!!
Thats a good place to stop🤩
A good bit of entertainment at 1am on my second beer. Funnily enough I clicked off of a technical explanation of polytonal lofi because I though this would be easier to grasp. Boy was I wrong. 😂
Adam Neely's video on p.l.?
I think you missed the cases where one of the variables is bigger than 2 and the other is not.
Well if you want to include it the 7^0, 7^1, 3^2 and 3^1 options were already considered and its fairly easy to notice that if the RHS is 3^0 there are no solutions and that 7^2 = 49 is not 2 less than a power of 3.
0 is a NATURAL NUMBER??
Some mathematicians agree, some disagree. It is still being debated.
@@angelmendez-rivera351 That is not the whole truth. In germany it is still convention that the set of natural numbers does not include 0. Also it is not about exluding 0 from a set, but excluding it from the particular set of the naturals. Of course them having a neutral element under addition might be handy, but often it isn't. That is why there is still debate.
Damn you've lost a lotta muscle. The previous vid of yours which i saw was from may where you look pretty jacked. Good problem btw.