$a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$ according to the identity.

WakeUp wrote: Find all positive integers $a,b$ for which $a^4+4b^4$ is a prime number. $a^4+4b^4=(a^2+2b^2-2ab)(a^2+2b^2+2ab)$ $\implies a^2+2b^2-2ab=1\implies (a-b)^2+b^2=1\implies (a,b)\equiv (1,0),\ (1,1)$ $\implies(a,b)\equiv (1,1)$

By the Sophie Germain factorization, we need $a^2-2ab+2b^2=1$, which rewrites as $(a-b)^2+b^2=1$. Then either $a-b=0,b=1$ giving $\boxed{(a,b)=(1,1)}\Rightarrow a^4+4b^4=5$, or $a-b=1,b=0$ which isn't a positive integer.

Seriously?

Well since this is bumped... The answer is $(a,b)=(1,1)$. By Sophie-Germain, $a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$. Since we want $a^4+4b^4$ to be prime, we must have either $a^2+2ab+2b^2$ or $a^2-2ab+2b^2$ equal to $1$. But since $a,b\in \mathbb{N}^+$, we must have $a^2-2ab+2b^2=(a-b)^2+b^2=1$. By the Trivial Inequality, each of $(a-b)^2,b^2$ are non-negative, so we must have $a-b=1,b=0$ or the other way around. Checking, we find that the only solution is $(a,b)=(1,1)$.

Well since this is bumped... The only solution is $(a,b) = (1,1)$, which fits since it would imply $a^4 + 4b^4 = 5$. Now we show it's the only solution. We have \[a^4 + 4b^4 = (a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab) \] Note that both factors are positive integers, so one of them must be $1$. Since $a^2 + 2b^2 - 2ab$ is the smaller factor, it must be equal to $1$. This implies $(a-b)^2 + b^2 = 1$. Since $(a-b)^2 \ge 0$, we have $b^2 \le 1$, so $b=1$. This gives $a=1$, as desired.

By Sophie German $a^4+4b^4=(a^2-2ab+2b^2)(a^2+2ab+2b^2)$, so $a^2-2ab+2b^2=1$, so $(a-b)^2+b^2=1$, so $\boxed{(1,1)}$ is the only solution.

$a^4+4b^4=(a^2-2ab+2b^2)(a^2+2ab+2b^2)$ its because $a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2=(a^2+2b^2)^2-(2ab)^2=(a^2-2ab+2b^2)(a^2+2ab+2b^2)$