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

1. Chipper10 Group Title

Can you explain the notation on the right?

2. LolWolf Group Title

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 Group Title

So 60 is a factor of xyz or a multiple?

4. LolWolf Group Title

60 is a factor of $$xyz$$.

5. Chipper10 Group Title

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

6. LolWolf Group Title

The correction is up.

7. experimentX Group Title

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 Group Title

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

9. mukushla Group Title

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 Group Title

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

11. experimentX Group Title

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 Group Title

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

13. mukushla Group Title

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

14. experimentX Group Title

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

15. experimentX Group Title

let's try induction.

16. mukushla Group Title

17. experimentX Group Title

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 Group Title

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

19. mukushla Group Title

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 Group Title

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

21. experimentX Group Title

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

22. experimentX Group Title

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

23. experimentX Group Title

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

24. mukushla Group Title

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 Group Title

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

26. Herp_Derp Group Title

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 Group Title

o.O

28. experimentX Group Title

what the hell is that ....

29. experimentX Group Title

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

30. Herp_Derp Group Title

@mukushla I copied it from your pdf

31. mukushla Group Title

lol

32. LolWolf Group Title

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 Group Title

4. extra points to whomever gets the reference

34. mukushla Group Title

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

35. LolWolf Group Title

Yep. Nice job!