, 1994

A prime $p$ divides $a b$ if and only if divides either $a$ or $b$. If $n=a^{2}+b^{2}$ is a composite then it has a prime divisor $p \leq \sqrt{n}$, and if $p$ divides $a$ it divides $b$ and vice-versa, which is not possible because $a$ and $b$ are coprime. Therefore $n$ is a prime.

Suppose without loss of generality that $a \geq b$ and consider $a-b$. Note that $a^{2}+b^{2}=(a-b)^{2}+2 a b$.

- If $a=b$ then $a=b=1$ because $a$ and $b$ are coprime. $n=2$ is a solution.

- If $a-b=1$ then $a$ and $b$ are coprime and $a^{2}+b^{2}=(a-b)^{2}+2 a b=2 a b+1=2 b(b+1)+1= 2 b^{2}+2 b+1$. So any prime factor of any number smaller than $\sqrt{2 b^{2}+2 b+1}$ is a divisor of $a b=b(b+1)$.

One can check that $b=1$ and $b=2$ yields the solutions $n=1^{2}+2^{2}=5$ (the only prime $p$ is 2) and $n=2^{2}+3^{2}=13$ (the only primes $p$ are 2 and 3). Suppose that $b>2$.

Consider, for instance, the prime factors of $b-1 \leq \sqrt{2 b^{2}+2 b+1}$, which is coprime with $b$. Any prime must then divide $a=b+1$. Then it divides $(b+1)-(b-1)=2$, that is, $b-1$ can only have 2 as a prime factor, that is, $b-1$ is a power of 2 , and since $b-1 \geq 2$, $b$ is odd.

Since $2 b^{2}+2 b+1-(b+2)^{2}=b^{2}-2 b-3=(b-3)(b+1) \geq 0$, we can also consider any prime divisor of $b+2$. Since $b$ is odd, $b$ and $b+2$ are also coprime, so any prime divisor of $b+2$ must divide $a=b+1$. But $b+1$ and $b+2$ are also coprime, so there can be no such primes. This is a contradiction, and $b \geq 3$ does not yield any solutions.

- If $a-b>1$, consider a prime divisor $p$ of $a-b=\sqrt{a^{2}-2 a b+b^{2}}<\sqrt{a^{2}+b^{2}}$. Since $p$ divides one of $a$ and $b, p$ divides both numbers (just add or subtract $a-b$ accordingly.) This is a contradiction.

Hence the only solutions are $n=2,5,13$.