Primes and Powers

According to the Sophie Germain Identity,

$\begin{aligned}a^4 + 4b^4 & = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\ & = (a^2 + 2b^2)^2 - (2ab)^2 \\ & = (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)\end{aligned}$

Thus, $a^4+4b^4$ can be factorized and there is only one ordered pair for which $a^4+4b^4$ which is $1,1$ .

$a^{4}+b^{4}=(a^{2}+2ab+2b^{2})(a^2-2ab+2b^{2})$

Since we are looking for primes, the lowest factor must be equal to 1, so:

$a^{2}-2ab+2b^{2}=1$

$(a-b)^{2}+b^{2}=1$

$(a-b)^{2}=(1+b)(1-b)$

From here we can tell that the absolute value of b must be less than 2 otherwise it would make the RHS negative, this is not possible when the LHS is square.

We can consider the case $b=\pm1$

$(a-b)^{2}=(1+1)(1-1)$

$(a-b)^{2}=0$

$a=b=1$

Where $a^{4}+4b^{4}=1+4=\boxed{5}$

Then if $b=0$

$(a-b)^{2}=1$

$a=b+1=2$

Where $a^{4}+4b^{4}=16+4=20$

So the only such pair $(a,b)$ is $(1,1)$ which gives the solution 5, which is prime.