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*Math challenge* Prove: \[ x^2+y^2=z^2\implies 60|xyz\\ x,y,z\in \mathbb{Z} \] Why IS this site so addicting?

Mathematics
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Can you explain the notation on the right?
Yep, \(a|b\) means \(a\) divides \(b\), or a more formal definition: \[ a, b\in \mathbb{Z}\\ a|b\implies\exists k\in\mathbb{Z}\text{ s.t. }ak=b \]
So 60 is a factor of xyz or a multiple?

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Other answers:

60 is a factor of \(xyz\).
Are we assuming x,y,z are whole number cause if not that expression is definitely not always true.
The correction is up.
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.
60 = 2^2*3*5 let m=2
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
it seems all multiples of n* 3,4,5 is solution of the given condition;
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
n=4, 2* 4 * 15* 17 <-- multiple of 60
exper u let n=1 and i think we must work on general form
n=5, 10*24*26 n=6 .... probably yes for n>1
let's try induction.
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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
one of m,n is even so \[4|2mn\]so\[4|abc\]we need to prove \(3|abc\) and \(5|abc\)
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 \(3|abc\)
m * (m+1)(m-1) for any value of m, this is divisible by 3
m * (m+1)(m-1)(m^2 + 1) ^ ^ ^ can be of the form 5n+2 or 5n+3
(5n+2)^2 +1 = .... 2^2 + 1 = 5 (5n + 3)^2 + 1 = .... 3^2 + 1 = 10 both cases divisible by 5
for all n>1, the Pythagorean triplet is divisible by 60 (probably)
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 \(5|abc\)
so \(3,4,5|abc\) it gives \(60|abc\).
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 unramifi ed 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. \(\mathbf{Q.E.D.}\)
o.O
what the hell is that ....
can't follow ... probably not my stuff.
lol
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!
4. extra points to whomever gets the reference
if both of \(m,n\) are odd\[m^2-n^2\equiv m^2+n^2\equiv0 \ \ \text{mod} \ 2\]
Yep. Nice job!

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