How can one find all prime numbers $p,q,$ and $r$ such that $$(p-q)^2+1=r\ ?$$
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Yuval is right, this question is very hard. Breaking it up into 3 cases:
Edit: As Zev pointed out, $r$ even $\implies r = 2$ because if $r > 2$ we have $r - 2 = (p - q - 1)(p - q + 1)$ is even thus both terms $(p - q - 1),(p - q + 1)$ are and so we can continue to factor out $2$, i.e. $2^n | r - 2$ for all $n > 1$, a clear contradiction.
The Hardy-Littlewood conjecture E says that there are infinitely many primes of the form $n^2+1$ (and gives an asymptotic density); de Polignac's conjecture says that there are infinitely many primes, indeed consecutive primes, with a specified even distance. So on these conjectures there are infinitely many $r$ such that $r-1$ is a perfect square, and for any such choice ($r\neq2$) there are infinitely many $p,q$ such that $(p-q)^2+1=r.$