Find all prime numbers$\displaystyle K$
and all positive integers $\displaystyle N$
such that
$\displaystyle K = N^4+4.$
The only solution is $\displaystyle N=1,K=5$.
$\displaystyle N$ clearly has to be odd. If $\displaystyle N>1$, then $\displaystyle N$ must also be divisible by 5; if it isn’t, $\displaystyle N^4+4$ would be divisible by 5 by Fermat’s little theorem.
Hence if $\displaystyle N>1$, let $\displaystyle N=5(2k-1)$, $\displaystyle k\ge1$.
Then $\displaystyle N^4+4=625(2k-1)^4+4=\left[4(5k-2)^2+1\right]\left[4(5k-3)^2+1\right]$ is a product of two integers both greater than 1 and hence cannot be prime.
I found a better solution.
If $\displaystyle N$ is even, then $\displaystyle N^4+4$ is an even number greater than 2, so it is not prime.
If $\displaystyle N$ is odd, let $\displaystyle N=2m+1$, $\displaystyle m\ge0$.
Then $\displaystyle N^4+4=(2m+1)^4+4=\left(4m^2+1\right)\left(4(m+1)^2 +1\right)$.
If $\displaystyle m>0$, both the factors are greater than 1 and so $\displaystyle N^4+4$ is not prime. Hence the only solution is when $\displaystyle m=0$, i.e. $\displaystyle N=1$.