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Pythagorean triples
07-24-2016, 11:12 AM
Post: #3
(2) $a^2+b^2=2c^2$
Sol. We have $a^2-c^2=c^2-b^2,\;(a+c)(a-c)=(c+b)(c-b)$
$\quad$and clearly $a\equiv b\equiv c\text{ (mod 2)}.$ So $\gcd(a-c,c-b) \equiv 0\text{ (mod 2)}$
$\quad$Clearly $|a|=|c|=|b|\in\mathbb{N}$ satisfies the equation. Assume $|a|\ne |c|$ thus
$\quad (a-c)(c-b)\ne 0,\;\exists s,u,v\in\mathbb{Z}\setminus\{0\}:\;a-c = 2su,\,c-b =2sv,$
$\quad\gcd(u,v)=1.\;\therefore (a+c)u=(c+b)v,\;a+c = 2tv,\,c+b=2tu,$
$\quad$for some $t\in\mathbb{Z}\setminus\{0\}.$ Thus $c = tv-su=sv+tu,\;(s+t)u=(t-s)v$
$\quad = kuv$ for some $k\in\mathbb{Z}\setminus\{0\}$ since $\gcd(u,v)=1.$
$\quad$Now $(t+s,t-s) = k(v,u)\implies (t,s)=\frac{k}{2}(v+u,v-u)$ and finally
$\quad (a,b,c) = \frac{k}{2}(2uv+v^2-u^2,2uv-v^2+u^2,v^2+u^2)$
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Pythagorean triples - elim - 09-22-2010, 08:55 AM
RE: Pythagorean triples - elim - 09-23-2010, 07:23 PM
(2) $a^2+b^2=2c^2$ - elim - 07-24-2016 11:12 AM

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