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This conjecture seems to be true:

If the product of two different primes is a sum of two squares, the so are the primes.

I've tested it for $pq<1000$, but would like to see a proof.

Edit: If $p\neq q$ are primes and $pq=a^2+b^2$, then there are integers $c,d$ such that $p=c^2+d^2$.

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    What does "the so are the primes." mean?Do you mean If $pq=a^2+b^2$ then $a,b$ are primes? – 2017-01-27

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This is true by fermat's sum of two squares theorem.

the theorem states that an integer $n$ is a sum of two squares if and only if every prime $p\equiv \bmod 3$ that divides has an even exponent.

Using this:

If $pq$ is a sum of two squares we cannot have $p\equiv 3$ or $q\equiv 3$.

So both $p$ and $q$ are sums of two squares.