Exercise 5.4.15 ( There exist no integers $m>0$ and $n>0$ such that $m^2 + n^2, m^2 -n^2$ are perfect squares)

Proof. We show here that there do not exist positive integers $m,n,x,y$ such that

Note that (1) is equivalent to

If (2) has some solution, let $(x,y,m,n)\in {({\mathbb{N}}^{\text{∗}})}^4$ be a solution of the system (2) such that $y$ is minimal.

By (2), $y$ and $x$ are both odd or both even. If $2\mid y$ and $2\mid x$, then $2^2\mid y^2+x^2=2m^2$, so $2\mid m^2$, $2\mid m$, and similarly $2\mid n$. Then $(x∕2,y∕2,m∕2,n∕2)$ is a solution of (2) in positive integers such that $y∕2<y$, which contradicts the minimality of $y$. Moreover $y\wedge x=1$. Indeed, if some prime number $p$ satisfies $p\mid y,p\mid x$, then $p\ne 2$, and $p\mid 2m^2$, thus $p\mid m$, and similarly $p\mid n$. Then $(x∕p,y∕p,m∕p,n∕p)$ is a solution of (10) which contradicts the minimality of $y$. Therefore $y\wedge x=1$. Using (2), this implies $m\wedge x=m\wedge y=n\wedge y=n\wedge y=1$.

Since $y$ and $x$ are odd and $y>x$ by (1), we may write

where $\alpha$ is a positive integer. Then the first equation of (2) gives

thus

From $x^2+2\mathit{\alpha x}+2{\alpha }^2=m^2$ and $x\wedge m=1$, we see that $\alpha \wedge m=1$. This shows that $(x+\alpha ,\alpha ,m)$ is a primitive Pythagorean triple. By Theorem 5.5, there are positive integers $r,s$ of opposite parity such that $r\wedge s=1$ and

In either cases we have

Then the second equation of (2) gives, after division by $2$,

Since $r,s$ are of opposite parity, $r^2-s^2$ is odd, and $r\wedge s=1$, thus $(4\mathit{rs})\wedge (r^2-s^2)=1$. By Lemma 5.4, $r,s$ and $r^2-s^2$ are perfect squares, say

where $u,v,w$ are positive integers. Then $w^2=u^4-v^4$, or equivalently

But we have proved in Problem 14 that this equation has no solution in positive integers. This contradiction shows that there exist no positive integers $m$ and $n$ such that $m^2+n^2$ and $m^2-n^2$ are both perfect squares. □