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

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Chipper10 Group TitleBest ResponseYou've already chosen the best response.0
Can you explain the notation on the right?
 2 years ago

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

Chipper10 Group TitleBest ResponseYou've already chosen the best response.0
So 60 is a factor of xyz or a multiple?
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
60 is a factor of \(xyz\).
 2 years ago

Chipper10 Group TitleBest 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.
 2 years ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
The correction is up.
 2 years ago

experimentX Group TitleBest 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.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
60 = 2^2*3*5 let m=2
 2 years ago

mukushla Group TitleBest 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
 2 years ago

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

experimentX Group TitleBest 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
 2 years ago

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

mukushla Group TitleBest ResponseYou've already chosen the best response.4
exper u let n=1 and i think we must work on general form
 2 years ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.0
let's try induction.
 2 years ago

experimentX Group TitleBest 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
 2 years ago

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

mukushla Group TitleBest 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 \(3abc\)
 2 years ago

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

experimentX Group TitleBest 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
 2 years ago

experimentX Group TitleBest 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
 2 years ago

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

mukushla Group TitleBest 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 \(5abc\)
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
so \(3,4,5abc\) it gives \(60abc\).
 2 years ago

Herp_Derp Group TitleBest 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. \(\mathbf{Q.E.D.}\)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
what the hell is that ....
 2 years ago

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

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
@mukushla I copied it from your pdf
 2 years ago

LolWolf Group TitleBest 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!
 2 years ago

zhess21 Group TitleBest ResponseYou've already chosen the best response.0
4. extra points to whomever gets the reference
 2 years ago

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

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
Yep. Nice job!
 2 years ago
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