## LolWolf 3 years ago *Math challenge* Prove: $x^2+y^2=z^2\implies 60|xyz\\ x,y,z\in \mathbb{Z}$ Why IS this site so addicting?

1. Chipper10

Can you explain the notation on the right?

2. LolWolf

Yep, $$a|b$$ means $$a$$ divides $$b$$, or a more formal definition: $a, b\in \mathbb{Z}\\ a|b\implies\exists k\in\mathbb{Z}\text{ s.t. }ak=b$

3. Chipper10

So 60 is a factor of xyz or a multiple?

4. LolWolf

60 is a factor of $$xyz$$.

5. Chipper10

Are we assuming x,y,z are whole number cause if not that expression is definitely not always true.

6. LolWolf

The correction is up.

7. experimentX

the pythagoras triplets are given by {2m, m^2-1, m^2+1} 2m * (m^2 - 1) * (m^2+1) = 60 * k where k is an integer.

8. experimentX

60 = 2^2*3*5 let m=2

9. mukushla

if $$a^2+b^2=c^2$$ then general form of primitive pyth triples is in the form$a=m^2-n^2$$b=2mn$$c=m^2+n^2$where $$\gcd(m,n)=1$$ and one of $$m,n$$ is even

10. experimentX

it seems all multiples of n* 3,4,5 is solution of the given condition;

11. experimentX

for n=3, 2*3*(3^2 - 1) * (3^2 +1) = 2 * 3 * 2^3 * 2*5 <-- a multiple of 60 so, n * 6,8,10 is also multiple of 60

12. experimentX

n=4, 2* 4 * 15* 17 <-- multiple of 60

13. mukushla

exper u let n=1 and i think we must work on general form

14. experimentX

n=5, 10*24*26 n=6 .... probably yes for n>1

15. experimentX

let's try induction.

16. mukushla

17. experimentX

2m * (m^2 - 1) * (m^2+1) = 60 k 2* m * (m+1)(m-1)(m^2 + 1) ^ ^ ^ ^ ^ 1 2 3 4 5 let's try that for m=7 2*7*8*6*50

18. mukushla

one of m,n is even so $4|2mn$so$4|abc$we need to prove $$3|abc$$ and $$5|abc$$

19. mukushla

if $$3|m$$ or $$3|n$$ then $$3|abc$$ . if not then$m^2 \equiv1 \ \ \text{mod} \ 3$$n^2 \equiv1 \ \ \text{mod} \ 3$and$m^2-n^2 \equiv0 \ \ \text{mod} \ 3$so $$3|abc$$

20. experimentX

m * (m+1)(m-1) for any value of m, this is divisible by 3

21. experimentX

m * (m+1)(m-1)(m^2 + 1) ^ ^ ^ can be of the form 5n+2 or 5n+3

22. experimentX

(5n+2)^2 +1 = .... 2^2 + 1 = 5 (5n + 3)^2 + 1 = .... 3^2 + 1 = 10 both cases divisible by 5

23. experimentX

for all n>1, the Pythagorean triplet is divisible by 60 (probably)

24. mukushla

if $$5|m$$ or $$5|n$$ then $$5|abc$$ . if not then$m^2 \equiv 1,4 \ \ \text{mod} \ 5$$n^2 \equiv 1,4 \ \ \text{mod} \ 5$now if $$m^2 \equiv n^2 \ \ \text{mod} \ 5$$ then $$5|m^2-n^2$$ and $$5|abc$$ otherwise $$m^2 \not\equiv n^2 \ \ \text{mod} \ 5$$ $$c=m^2+n^2 \equiv1+4\equiv0 \ \ \text{mod}\ 5$$ so $$5|abc$$

25. mukushla

so $$3,4,5|abc$$ it gives $$60|abc$$.

26. Herp_Derp

FERMAT'S LAST THEOREM: Let $$n, a, b, c \in \mathbb{Z}$$ with $$n > 2$$. If $$a^n +b^n = c^n$$ then $$abc= 0$$. Proof: The proof follows a program formulated around 1985 by Frey and Serre. By classical results of Fermat, Euler, Dirichlet, Legendre, and Lamé, we may assume $$n = p$$, an odd prime $$>11$$. Suppose $$a, b, c \in \mathbb{Z}$$, $$abc \ne 0$$, and $$a^p + b^p = c^p$$. Without loss of generality we may assume $$2|a$$ and $$b \equiv 1 \mod 4$$. Frey observed that the elliptic curve $$E : y^2 = x(x - ap)(x + bp)$$ has the following "remarkable" properties: (1) $$E$$ is semistable with conductor $$N_E =\Pi_{\ell|abc}\ell$$; and (2) $$\bar{\rho}_{E,p}$$ is unramified outside 2p and is flat at at p. By the modularity theorem of Wiles and Taylor-Wiles, there is an eigenform $$f \in \mathcal{S}_2(\Gamma_0(N_E))$$ such that $$\rho_{f,p} =\rho_{E,p}$$. A theorem of Mazur implies that $$\bar{\rho}_{E,p}$$ is irreducible, so Ribet's theorem produces a Hecke eigenform $$g \in \mathcal{S}_2(\Gamma_0(2))$$ such that $$\rho_{g,p} \equiv \rho_{f,p} \mod\wp$$ for some $$\wp$$. But $$X_0(2)$$ has genus zero, so $$\mathcal{S}_2(\Gamma_0(2)) = 0$$. This is a contradiction and Fermat's Last Theorem follows. $$\mathbf{Q.E.D.}$$

27. mukushla

o.O

28. experimentX

what the hell is that ....

29. experimentX

can't follow ... probably not my stuff.

30. Herp_Derp

@mukushla I copied it from your pdf

31. mukushla

lol

32. LolWolf

Fermat's? Lol, only like 100 people in the world understand Wiles' proof of the Taniyama–Shimura conjecture XD But, anyways, you guys are correct, but missing one essential step: Prove that if both of $$m,n$$ are odd, then it is not a primitive solution. Otherwise, you've got it, nice job!

33. zhess21

4. extra points to whomever gets the reference

34. mukushla

if both of $$m,n$$ are odd$m^2-n^2\equiv m^2+n^2\equiv0 \ \ \text{mod} \ 2$

35. LolWolf

Yep. Nice job!