## anonymous 4 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. anonymous

Can you explain the notation on the right?

2. anonymous

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. anonymous

So 60 is a factor of xyz or a multiple?

4. anonymous

60 is a factor of $$xyz$$.

5. anonymous

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

6. anonymous

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. anonymous

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. anonymous

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. anonymous

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. anonymous

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

19. anonymous

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. anonymous

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. anonymous

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

26. anonymous

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. anonymous

o.O

28. experimentX

what the hell is that ....

29. experimentX

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

30. anonymous

@mukushla I copied it from your pdf

31. anonymous

lol

32. anonymous

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. anonymous

4. extra points to whomever gets the reference

34. anonymous

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

35. anonymous

Yep. Nice job!