Exercise 1.23

Proof. Suppose $x^2+y^2=z^2,x,y,z\in \mathbb{Z}$. Let $d=x\wedge y\wedge z$. If $d=0$, then $x=y=z=0$. If $d\ne 0$, and $a=x∕d,b=y∕d,c=z∕d$, then $a^2+b^2=c^2$, with $a\wedge b\wedge c=1$. If a prime $p$ is such that $p\mid a,p\mid b$, then $p\mid c^2$, so $p\mid c$ (as $p$ is a prime). Then $p\mid a\wedge b\wedge c=1$ : this is impossible, so $a\wedge b=1$, and similarly $a\wedge c=1,b\wedge c=1$.

If $a,b$ are odd, then $a^2\equiv b^2\equiv 1(\mathit{mod}4)$, so $c^2\equiv 2(\mathit{mod}4)$. As the squares modulo 4 are $0,1$, this is impossible. As $a\wedge b=1$, $a,b$ are not both even, so $a,b$ are not of the same parity. Without loss of generality, we may exchange $a,b$ so that $a$ is even, $b$ is odd, and then $c$ is odd.

$a^2=c^2-b^2=(c-b)(c+b)$, so

where $a∕2,(c-b)∕2,(c+b)∕2$ are integers.

If $d\mid (c-b)∕2$ and $d\mid (c+b)∕2$, then $d\mid c=(c+b)∕2+(c-b)∕2$, and $d\mid b=(c-b)∕2-(c-b)∕2$, so $d\mid c\wedge b=1$. This proves

Using Ex. 1.16, we see that $(c+b)∕2$ and $(c-b)∕2$ are squares : there exist $u,v$ such that

${(a∕2)}^2=u^2{v}^2$, and we can choose the signs of $u,v$ such that $a=2\mathit{uv}$. Then $b=v^2-u^2,c=v^2+u^2$. There exists $\lambda \in \mathbb{Z}$ ( $\lambda =d$) such that $x=2\mathit{\lambda uv},y=\lambda (v^2-u^2),z=\lambda (v^2+u^2)$.

Conversely, if $\lambda ,u,v$ are any integers, ${(2\mathit{\lambda uv})}^2+(\lambda {(v^2-u^2)}^2={\lambda }^2(4u^2{v}^2+v^4+u^4-2u^2{v}^2)={\lambda }^2(v^4+u^4+2u^2{v}^2)={(\lambda (u^2+v^2))}^2$.

Conclusion : if $x,y,z\in \mathbb{Z}$,

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