Find the natural numbers such that a number is a prime number.

Question

Find the natural numbers $x$, $n$ such that $p = x^4+2^{4n+2}$ is prime number.

Answer 1

Rewrite $p$ as follows, $p=x^4+2^{4n+2}=x^4+4^{2n+1}=x^4+4\cdot 4^{2n}$. Now, $$x^4+4y^4=(x^2)^2+(2y^2)^2+2\cdot x^2\cdot 2y^2-4x^2y^2=(x^2+2y^2)^2-(2xy)^2\\=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$$

We need $x^2-2xy+2y^2=1$, as this is the minimum factor, to make $p$ a prime. Here we have $y=4^n$. Someone already mentioned that $p=5$ is a solution. In fact it is the only solution.

Can you continue from here?