BMO Short List

The required numbers are all even positive integers alone. Indeed, if $n$ is even, then let $a=n^2/2-1$ and $b=n^2/2+1$, to check that $\frac{a^2+n^2}{b^2-n^2}=\frac{n^4/4+1}{n^4/4+1}=1$.

Suppose now that such $a$ and $b$ exist for some positive odd integer $n$. Notice that we may and will assume $\text{gcd}(n,a,b)=1$. Note also that $n^2\equiv 1(\mathrm{mod}4)$. If $b$ is odd, then $b^2-n^2$ is divisible by $4$, and hence so is $a^2+n^2$. Since $n$ is odd, $a^2+1$ is therefore divisible by $4$, which is impossible. Thus, $b$ must be even. Then $b^2-n^2\equiv 3(\mathrm{mod}4)$, so $b^2-n^2$ has a prime factor $p\equiv 3(\mathrm{mod}4)$. Then $a^2+n^2$ is divisible by $p$, and it follows that so are both $a$ and $n$, since $p\equiv 3(\mathrm{mod}4)$. On the other hand, since $n$ and $b^2-n^2$ are both divisible by $p$, so is $b$. Consequently, $a$, $b$ and $n$ are all divisible by $p$, contradicting the assumption $\text{gcd}(n,a,b)=1$.