Sum of 2 Squares in 2 Distinct Ways

Theorem
Let $m, n \in \Z_{>0}$ be distinct positive integers that can be expressed as the sum of two distinct square numbers.

Then $m n$ can be expressed as the sum of two square numbers in at least two distinct ways.

Proof
Let:
 * $m = a^2 + b^2$
 * $n = c^2 + d^2$

Then:

It remains to be shown that $a \ne b$ and $c \ne d$, then the four numbers:


 * $a c + b d, a d - b c, a c - b d, a d + b c$

are distinct.

Since $a,b,c,d > 0$, $a c + b d \ne a c - b d$ and $a d + b c \ne a d - b c$.

We also have:

Thus $a \ne b$ and $c \ne d$ implies $a c \pm b d \ne a d \pm b c$.

The case for $a c \pm b d \ne a d \mp b c$ is similar.