A throwback number theory problem

[@jkid1134]

Better than being a throwback to the Balkans in the 90s

[@goodplacetostop2973]

11:01 Spoiler alert, the original statement says to… Prove that the equation y^2 = x^5 - 4 has no integer solutions. So yeah, no solution was the answer

[@Bodyknock]

One small thing that makes it slightly easier at 8:00 is, instead of doing the chart from n=0 to 10, do it from n= -5 to 5. You get the same results but it's immediately obvious why there's symmetry about 0 and why you only need to actually calculate n=0 through 5.

[@aln4075]

So satisfying to watch an olympiad problem being solved👍🏻

[@wojteksocha2002]

If p=4k+3 and p divides x^2 + y^2, then p divides both x and y. That means that 4 divides both sides and 8 does not, but if 4 divides y^5, then 32 divides y^5

[@FiniteSimpleFox]

I'm sorry but I don't quite understand your solution. I see why the first line is true due to the sum of two squares thrm, but it's unclear to me what exactly you're talking about when you say "that means that 4 divides both sides and 8 does not", or how you got to that point.

[@virajagr]

6:11 ah yes, 10 = 1 -11, the good old Michael is back

[@HershO.]

This was a nice revision of number theory.

[@sonhoangngan4088]

That's great. Tks much😊

[@jesusalej1]

According to FLT if y^5 = 1(mod 3), then x^2+1 must be 1(mod 3), that means x^2 = 0(mod 3). From there come the solutions, x^2+1 cannot be 2(mod 3), because is a contradiction.

[@RexxSchneider]

I don't see a contradiction. We know that x^2+1 ≡ {1, 2, 2} (mod 3) depending on whether x ≡ {0, 1, 2} (mod 3). So the RHS can be congruent to either 1 or 2. But we find that y^5 ≡ {0, 1, 2} (mod 3) where y ≡ {0, 1, 2} (mod 3). So the LHS can be 0 or 1 or 2. The only possibility you can eliminate from that is that y is a multiple of 3. Why can't x^2+1 ≡ 2 (mod 3)? Any value of x that isn't a multiple of 3 will satisfy that, and any value of y ≡ 2 (mod 3) will make y^5 ≡ 2 (mod 3).

[@jesusalej1]

@@RexxSchneider u r right man, bad thinking. Greetings.

[@mehmetozdemir8115]

Why are mod3 and mod11 enough to say there is no solution?

[@karl131058]

Any polynomial equation containing only integers must remain true if you take both sides mod . So, if there WERE any solutions, they would also be solutions mod 11 (for example 😉). Since there aren't any mod 11, there aren't any at all!

[@rickdesper]

Mod 11 is enough to say there is no solution. He used mod 3 just to see if he could better understand the problem and possibly find some solutions. The proof is, to summarize, that if x and y are integral solutions to y^5 = x^2 + 4, then x^2 must be congruent to either 6 or 8 mod 11, but that's impossible.

[@dandjr1546]

So he checked mod 3 and mod 11. Does that prove there are no solutions? Or do you also have to check mod 17, mod 31, .. mod 1187, ...

[@lagnugg]

yes, checking only mod 11 is enough for this problem. the big idea is that we have to have x² = a (mod 11), but x² doesn't cover all numbers mod 11 (or mod any prime), so if we can not have x² = a (mod 11), then there is no solution in terms of x

[@leif1075]

Surely you can solve without fermats little theorem or modular arithmetic so why use them at all?

[@RexxSchneider]

Because using Fermat's Little Theorem allows you to eliminate one of the variables, which then shows that no solutions are available. Please feel free to sketch out your alternative proof that doesn't use modular arithmetic.

[@leif1075]

@@RexxSchneider right but if you dont think of fermats theorem and im.sure a lot of ppl dont then surely there must be another way?

[@RexxSchneider]

@@leif1075 Sorry, but I don't know of another methodical way to show that no solutions exist. Perhaps a knowledge of Fermat's Little Theorem is necessary in order to solve the problem?

[@rickdesper]

Have at it! Better proofs are always welcome.

[@dkravitz78]

So technically you could have just made a chart for all fifth powers of y mod 11, You would immediately see that it is always 1 or 10, or 0 if 11 dovides y. So then you'd subtract 4 and say x^2 mod 11 is 7,6,8, and none of them are squares. Easier with FLT but not necessary

[@Anonymous-zp4hb]

Ah quadratic residues. Of course. Didn't get very far with this problem. Nice solution.

[@АнтонФ-ц5й]

Hi. 10!/6! = 7! Any other a,b,c from N (Z) exsists?

[@georgelaing2578]

Another first-rate video!

[@lawrencecataylo7712]

So if the value of y^5 = x^2 + 4 = 1(mod 3) and y^10 = (x^2 + 4)^2 (mod 11), how do we get the modulo value, based on Fermat's little theorem? That is the modulo 3 and 11? is it like the main remainder of the given equation of y? thank you.

[@RexxSchneider]

Fermat's little theorem states that a^(p-1) ≡ 1 (mod p) if p does not divide a. We want to use that to eliminate either x or y from the equation y^5 = x^2 + 4. We can eliminate x by treating x as a and (p-1) as 2. That means we reduce mod 3. So we have x^2 (mod 3) ≡ 1 if x is not divisible by 3 and that lets us reduce the original equation to y^5 ≡ 1 + 4 ≡ 2 (mod 3). That is the motivation for reducing mod 3. Alternatively, we could try to eliminate y from the original equation, but if y^5 is the same as a^(p-1) then p would have to be 6, which isn't a prime. However, if we square y^5, we get y^10, and treating that as a^(p-1) would imply p=11, which is a prime. That means the original equation squared would be y^10 = (x^2+4)^2 and we can reduce that mod 11 to eliminate y, giving 1 ≡ (x^2+4)^2 (mod 11) in the case where x is not divisible by 11. Does that now explain any better why we could choose mod 3 to eliminate x or mod 11 to eliminate y from the original equation?

[@johns.8246]

How about for x^2=y^5+4 ? I already see a solution!

[@atikshagarwal5147]

6,2 is asoln

[@joaopedrogoncalvesramos2218]

Factor (x-2)(x+2) = y^5. As gcd(x-2,x+2) | 4, we have two cases: 1) y odd. Then x is odd, and the gcd above is one. Hence both expressions are fifth powers, that is: x-2=a^5, x+2=b^5. Subtract one from the other to see that 4 = b^5-a^5. Since fifth powers grow fast, it’s easy to check this is never 4. Hence we are led to the othe rcase: 2) y even. Then x is even, and the gcd before is either 2 or 4. A quick check and, since both factors are congruent mod 4, the gcd has to be 4. Hence, if x=4k+2, y = 2m, (k+1)k = 2m^5. Now we play the same game: the gcd of both factors is 1, so one of them is a fifth power and the other is twice a fifth power. Case 2.1) k+1 = p^5, k=2q^5. In this case, we have p^5 -2q^5 = 1. Then the problem gets interesting; in order to solve this, it seems we have to work on Z[2^{1/5}]. There, we want to show that the only guy of the form p - q 2^{1/5} whose norm in that ring is \pm 1 is with p=q=-1. For the Case 2.2 we obtain similarly 2p^5 - q^5 = 1. The same analysis concludes, although I’m too lazy to carry it out… Alternatively, one may use the condition on p,q to write |2^{1/5} - p/q| < C/q^5, for instance in Case 2.1. But since all algebraic numbers are approximable to order at most 2, this equation has finitely many solutions

[@quite_unknown_1]

Mod 11 is sufficiënt on itself, no need to check mod 3

[@RexxSchneider]

That's true, but why would anybody immediately decide to check mod 11? The obvious application of FLT is to check mod 3, and only when that is inconclusive does it make sense to look for a larger prime to use for the modulo check.

[@de_michael1222]

@@RexxSchneider there is a theorem that if p is a prime other than 2, then for all x s.t. p doesnt divide x, x^((p-1)/2) = +/- 1 (mod p) which gives motivation for checking mod 11

[@RexxSchneider]

@@de_michael1222 Yes, it's just a corollary of Fermat's Little Theorem that I mentioned in my previous post. Since x^(p-1) ≡ 1 (mod p), it should be immediately obvious that taking the square root of each side gives you x^((p-1)/2) ≡ ±1 (mod p). I understood that would be the motivation for picking 11. But you still haven't explained why you would choose to check mod 11 _before_ checking the mod 3 case? Surely it's sensible to do the most straightforward case first? If you use Euler's Theorem, you know that a^φ(n) ≡ 1 (mod n) for any n, as long as gcd(a, n) = 1, so you could potentially reduce the equation modulo any number you choose, but it's still sensible to examine the easiest cases first.

[@de_michael1222]

@@RexxSchneider regarding mod 3 case, I briefly checked it in my head but didn't go that way bc it was easy to see after a couple of seconds that it won't help. But yeah you're kinda right, i didn't move straight to checking mod 11

[@quite_unknown_1]

@@RexxSchneider I'm not trying to flex here at all but my first instinct is to check mod 11 in these type of questions; like actually 10 seconds and I got the solution. It's a really common technique tbf, just modulo checking 2,3,5,7,11 can solve like 10% of all the NT questions.

[@frentz7]

You put a *hint* on the chalkboard *with* the statement of the problem. Come on .... at least put the problem up there for five seconds, for people who want to try it!

[@frentz7]

I mean (sorry) keep the great videos coming, too. :) But no hints?

[@letswait30days]

Then just look at the thumbnail and try it from there

[@ayoubabid213]

Its an easy problem

[@petersievert6830]

difficulty is always a matter of perspective. good on you, if you are on a level to easily handle these tasks.