Question

*Math challenge*
Prove:
\[
x^2+y^2=z^2\implies 60|xyz\\
x,y,z\in \mathbb{Z}
\]

Why IS this site so addicting?

Answer 1

Can you explain the notation on the right?

Answer 2

Yep, \(a|b\) means \(a\) divides \(b\), or a more formal definition:
\[
a, b\in \mathbb{Z}\\
a|b\implies\exists k\in\mathbb{Z}\text{ s.t. }ak=b
\]

Answer 3

So 60 is a factor of xyz or a multiple?

Answer 4

60 is a factor of \(xyz\).

Answer 5

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

Answer 6

The correction is up.

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

let's try induction.

Answer 16

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

Answer 17

one of m,n is even so \[4|2mn\]so\[4|abc\]we need to prove \(3|abc\) and \(5|abc\)

Answer 18

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 \(3|abc\)

Answer 19

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

Answer 20

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

Answer 21

(5n+2)^2 +1 = .... 2^2 + 1 = 5
(5n + 3)^2 + 1 = .... 3^2 + 1 = 10

both cases divisible by 5

Answer 22

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

Answer 23

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 \(5|abc\)

Answer 24

so \(3,4,5|abc\) it gives \(60|abc\).

Answer 25

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.
\(\mathbf{Q.E.D.}\)

Answer 26

o.O

Answer 27

what the hell is that ....

Answer 28

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

Answer 29

@mukushla I copied it from your pdf

Answer 30

lol

Answer 31

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!

Answer 32

4. extra points to whomever gets the reference

Answer 33

if both of \(m,n\) are odd\[m^2-n^2\equiv m^2+n^2\equiv0 \ \ \text{mod} \ 2\]

Answer 34

Yep. Nice job!