Are there infinitely many primitive Pythagorean triples for which 2 of the sides are prime? I use $a=m^{2}-n^{2}=(m-n)(m+n)$, $b=2mn$, and $c=m^{2}+n^{2}$. So I can start on making $a$ prime by letting $m=n+1$, giving $m^2-n^2=(m-n)(m+n)=2n+1$, and then choosing values of $n$ so that $2n+1$ is prime. Then $c=m^{2}+n^{2}=(n+1)^{2}+n^{2}=2(n^{2}+n)+1$, and we similarly need values of $n$ so that $2(n^{2}+n)+1$ is prime. I can imagine there being infinitely many values of $n$ that make each expression prime ($c$ being less obvious), but I am unsure how one could conclude that there are (or there are not) infinitely many $n$ for which both $a$ and $c$ are prime.

Since $$a^2+b^2=c^2\iff\begin{cases}a=m^2-n^2\\b=2mn\\c=m^2+n^2\end{cases}$$ the only candidates to be primes are $a$ and $c$ which happen only when $m=n+1$. It follows the equation in primes $p$ and $q$ $$p^2+1=2q$$The proposed question leads to know if this last equation has infinitely many solutions which is a variant of the still non proved conjecture of Bunyakovsky generalizing the Dirichlet's theorem on the infinitude of primes in arithmetical progressions.This is a very hard question and for the second degree you can see at the paper of
Betty Garrison: Polynomials With Large Numbers of Prime Values." The American Mathematical Monthly, 97(4), pp. 316–317 in which is studied the number possible of primes for irreducible polynomials of degree two (but not for $p^2+1$ but for $x^2+1$ and not giving twice a prime, $2q$ but a prime $q$ so the proposed problem here  it's even more difficult).
