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

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Chipper10Best ResponseYou've already chosen the best response.0
Can you explain the notation on the right?
 one year ago

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

Chipper10Best ResponseYou've already chosen the best response.0
So 60 is a factor of xyz or a multiple?
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
60 is a factor of \(xyz\).
 one year ago

Chipper10Best ResponseYou've already chosen the best response.0
Are we assuming x,y,z are whole number cause if not that expression is definitely not always true.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
the 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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
60 = 2^2*3*5 let m=2
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
if \(a^2+b^2=c^2\) 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
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
n=4, 2* 4 * 15* 17 < multiple of 60
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
exper u let n=1 and i think we must work on general form
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
n=5, 10*24*26 n=6 .... probably yes for n>1
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
let's try induction.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
2m * (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
 one year ago

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

mukushlaBest ResponseYou've already chosen the best response.4
if \(3m\) 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 \(3abc\)
 one year ago

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

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

experimentXBest 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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
for all n>1, the Pythagorean triplet is divisible by 60 (probably)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
if \(5m\) 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 \(5abc\)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
so \(3,4,5abc\) it gives \(60abc\).
 one year ago

Herp_DerpBest ResponseYou've already chosen the best response.0
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 \(2a\) 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. \(\mathbf{Q.E.D.}\)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
what the hell is that ....
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
can't follow ... probably not my stuff.
 one year ago

Herp_DerpBest ResponseYou've already chosen the best response.0
@mukushla I copied it from your pdf
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
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!
 one year ago

zhess21Best ResponseYou've already chosen the best response.0
4. extra points to whomever gets the reference
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
if both of \(m,n\) are odd\[m^2n^2\equiv m^2+n^2\equiv0 \ \ \text{mod} \ 2\]
 one year ago
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