# proof of Pythagorean triples

Suppose that $a^{2}+b^{2}=1$ where $a,b\in\mathbb{Q}$. $a^{2}+b^{2}=N_{\mathbb{Q}(i)/\mathbb{Q}}(a+bi)$ (here $\operatorname{N}$ is the norm), so $a^{2}+b^{2}=1$ if and only if $N_{\mathbb{Q}(i)/\mathbb{Q}}(a+bi)=1$. $\mathbb{Q}(i)$ is cyclic over $\mathbb{Q}$ with Galois group isomorphic to $\mathbb{Z}/2\mathbb{Z}$, so by Hilbert’s Theorem 90, there is some element $s+ti\in\mathbb{Q}(i)$ such that

 $a+bi=\frac{s+ti}{\sigma(s+ti)}=\frac{s+ti}{s-ti}=\frac{s^{2}-t^{2}+2sti}{s^{2}% +t^{2}}$

so that

 $a=\frac{s^{2}-t^{2}}{s^{2}+t^{2}},\qquad b=\frac{2st}{s^{2}+t^{2}}$

Now, given any integer right triangle $p,q,r$ with $p^{2}+q^{2}=r^{2}$, we have

 $\left(\frac{p}{r}\right)^{2}+\left(\frac{q}{r}\right)^{2}=1$

where $p/r,q/r\in\mathbb{Q}$, so for some $s,t\in\mathbb{Q}$,

 $\frac{p}{r}=\frac{s^{2}-t^{2}}{s^{2}+t^{2}},\qquad\frac{q}{r}=\frac{2st}{s^{2}% +t^{2}}$

Clearing fractions on the right hand side of these equations by multiplying numerator and denominator by the square of the least common multiple of the denominators of $s,t$, we get

 $\frac{p}{r}=\frac{m^{2}-n^{2}}{m^{2}+n^{2}},\qquad\frac{q}{r}=\frac{2mn}{m^{2}% +n^{2}}$

for $m,n\in\mathbb{Z}$. Thus for some $d\in\mathbb{Z}$,

 $p=d(m^{2}-n^{2}),\qquad q=2mnd,\qquad r=d(m^{2}+n^{2})$
Title proof of Pythagorean triples ProofOfPythagoreanTriples1 2013-03-22 17:44:34 2013-03-22 17:44:34 rm50 (10146) rm50 (10146) 5 rm50 (10146) Proof msc 11-00