Pick any solution of $a^2 + b^2 = c^2$, such that $2|b$, but $2 \nmid a$ and $2 \nmid c$. Then $b^2$ is an amiable number. Indeed, we can partition $ \{1, 2, ..., b^2 \}$ into pairs $$ \{1, a^2 - 1 \}, \{2, a^2 - 2 \}, ..., \{ \frac{a^2-1}{2}, \frac{a^2+1}{2} \} $$ and $$ \{a^2, b^2 \}, \{a^2+1, b^2-1 \}, ..., \{ \frac{a^2+b^2 - 1}{2}, \frac{a^2+b^2 + 1}{2} \} $$