Are there an infinite number of primes satisfying these three conditions? (1)   $p$ is prime, and so is $(p+1)/2$.
(2)   $p$ can be written as $a^2 + b^4$, $a,b \in \mathbb{N}$.
(3)   $p \equiv 2 \bmod 31$.
These three conditions were contrived to make a $2017$ puzzle. Now I ask the title
question just out of curiosity.
The second condition means that $p$ is a Friedlander–Iwaniec prime.
It is known there are an infinite number of these primes.
No doubt it would be challenging to extend their theorem, but possibly
there are only a finite number satisfying all three conditions.
John Chessant calculated out to the prime $4914277 = 186^2 + 47^4$.
Perhaps much easier would be to show there are an infinite number satisfying (1) & (3).
 A: I'd be suprised if even the infinitude of (1) was known, so I think this is an extremely hard problem. As I said in a comment, a related problem, that both $p$ and $(p-1)/2$ are prime, is equivalent to the infinitude of Sophie Germain primes, and a lot of research has been done towards that problem. I would suspect that we could find it if the infinitude of (1) was known. 
However, there are infinitely many such primes if Schinzel's hypothesis H is true. 

Then, take $f(x)=(31(2x-1)+1)^2+1^4$ and $g(x)=\frac12\left((31(2x-1)+1)^2+2\right)=\frac12(f(x)+1)$. 
We need to prove that for every prime $p$, there is an $x$ such that $p \nmid f(x)g(x)$. 
Let $p$ be odd. 
Now if $x=\frac{p+1}{2}$, then we have $2x-1=p$, hence $$f(x)=(31(2x-1)+1)^2+1^4 \equiv 1^2+1^4 = 2 \mod p$$
$$g(x)= \frac12(f(x)+1) \equiv \frac{3p+3}{2}  \mod p$$
If $p \neq 2$, we have $p \nmid f(x)$, $p \nmid g(x)$ and hence $p \nmid f(x)g(x)$.If $p=2$, note that $f(x) \equiv 1 \mod 4$ and hence $g(x) \equiv 1 \mod 4$ for all $x$, so then also $p \nmid f(x)g(x)$.
Therefore, if Schinzel's hypothesis H is true, there are infinitely many $x$ such that both $f(x)=(31(2x-1)+1)^2+1^4$ and $g(x)=\frac12(f(x)+1)$ are prime.
Now, we check that $f(x)$ satisfies the properties:


*

*$f(x)$ and $\frac12(f(x)+1)$ are prime.

*$f(x)=  a^2 + b^4$ with $a=62x-1$ and $b=1$. 

*$f(x) \equiv 1^2 + 1^4=2 \mod 31$. 


Therefore, $f(x)$ satisfies.
