Pythagorean triples

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(1) Find all $(a,b,c) \in \mathbb{N}^+$ such that $a^2+b^2=c^2$
(2) Find all $(a,b,c) \in \mathbb{N}^+$ such that $a^2+b^2=2 c^2$

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A positive integer triple $(a,b,c)$ is called Pythagorean if $a^2+b^2=c^2 $
A Pythagorean triple $(a,b,c)$ is prime if $gcd(a,b,c)=1$
(1) Let $(a,b,c)$ be a prime Pythagorean triple, then $c \equiv a-b \equiv 1 \mod {2}$
$\quad\quad$Since $\gcd(a,b)>1 \Rightarrow \gcd(a,b,c)>1$ and the sum of two odd squares is not a square
$\quad\quad$Assume $2\mid b$ we then have $2 \nmid a, \; 2 \mid (c \pm a)$ and $\displaystyle{\frac{b^2}{4} = \frac{c+a}{2}\frac{c-a}{2}}$ a perfect square. Since
$\quad\quad \displaystyle{\gcd(\frac{c+a}{2},\frac{c-a}{2})=\gcd(c,\frac{c-a}{2})}\le \gcd(c,c-a)=\gcd(a,c)=1$, we see that
$\quad\quad \displaystyle{\frac{c-a}{2} = m^2, \; \frac{c+a}{2} = n^2, \; \gcd(m,n)=1}$ Therefore the set of prime Pythagorean triples is
$\quad\quad P_0 =\{(a,b,c) | c = m^2+n^2,\; \{a,b\} = \{n^2-m^2,2mn\}, \; m<n,\; 2 \nmid (n-m), \; \gcd(m,n) = 1\}$
$\quad\quad$ While the set of Pythagorean triples is
$\quad\quad P_* = \{(a,b,c) | c = m^2+n^2, \; \{a,b\} = \{n^2-m^2,2mn\}, \; m,n \in \mathbb{N}^+,\; m<n\} $

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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
$\quad (a,b,c) = \frac{k}{2}(2uv+v^2-u^2,2uv-v^2+u^2,v^2+u^2)$