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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you explain the notation on the right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yep, \(ab\) means \(a\) divides \(b\), or a more formal definition: \[ a, b\in \mathbb{Z}\\ ab\implies\exists k\in\mathbb{Z}\text{ s.t. }ak=b \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 60 is a factor of xyz or a multiple?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.060 is a factor of \(xyz\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are we assuming x,y,z are whole number cause if not that expression is definitely not always true.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The correction is up.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0the pythagoras triplets are given by {2m, m^21, m^2+1} 2m * (m^2  1) * (m^2+1) = 60 * k where k is an integer.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.060 = 2^2*3*5 let m=2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if \(a^2+b^2=c^2\) then general form of primitive pyth triples is in the form\[a=m^2n^2\]\[b=2mn\]\[c=m^2+n^2\]where \(\gcd(m,n)=1\) and one of \(m,n\) is even

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0it seems all multiples of n* 3,4,5 is solution of the given condition;

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0for 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0n=4, 2* 4 * 15* 17 < multiple of 60

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0exper u let n=1 and i think we must work on general form

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0n=5, 10*24*26 n=6 .... probably yes for n>1

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0let's try induction.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.02m * (m^2  1) * (m^2+1) = 60 k 2* m * (m+1)(m1)(m^2 + 1) ^ ^ ^ ^ ^ 1 2 3 4 5 let's try that for m=7 2*7*8*6*50

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0one of m,n is even so \[42mn\]so\[4abc\]we need to prove \(3abc\) and \(5abc\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if \(3m\) or \(3n\) then \(3abc\) . if not then\[m^2 \equiv1 \ \ \text{mod} \ 3\]\[n^2 \equiv1 \ \ \text{mod} \ 3\]and\[m^2n^2 \equiv0 \ \ \text{mod} \ 3\]so \(3abc\)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0m * (m+1)(m1) for any value of m, this is divisible by 3

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0m * (m+1)(m1)(m^2 + 1) ^ ^ ^ can be of the form 5n+2 or 5n+3

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0(5n+2)^2 +1 = .... 2^2 + 1 = 5 (5n + 3)^2 + 1 = .... 3^2 + 1 = 10 both cases divisible by 5

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0for all n>1, the Pythagorean triplet is divisible by 60 (probably)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if \(5m\) or \(5n\) then \(5abc\) . 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 \(5m^2n^2\) and \(5abc\) otherwise \(m^2 \not\equiv n^2 \ \ \text{mod} \ 5\) \(c=m^2+n^2 \equiv1+4\equiv0 \ \ \text{mod}\ 5 \) so \(5abc\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so \(3,4,5abc\) it gives \(60abc\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0FERMAT'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 \(2a\) 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 \(N_E =\Pi_{\ellabc}\ell\); and (2) \(\bar{\rho}_{E,p}\) is unramified outside 2p and is flat at at p. By the modularity theorem of Wiles and TaylorWiles, 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. \(\mathbf{Q.E.D.}\)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0what the hell is that ....

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0can't follow ... probably not my stuff.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mukushla I copied it from your pdf

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Fermat'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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.04. extra points to whomever gets the reference

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if both of \(m,n\) are odd\[m^2n^2\equiv m^2+n^2\equiv0 \ \ \text{mod} \ 2\]
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