# primes and the sum of two squares

I got the following problem.

Prove that $p=a^4+4b^4 \;\;\;$can not be a prime number, $\;\; p\gt 5$

I know that $p=(a^2)^2+(2b^2)^2$ can be written as the sum of two squares if and only if $\;\;p\equiv 1 \mod \;4$.
Hence $a \equiv 1$ or $a\equiv 3 \mod \;4$

I dont know how can I continue from here

• Why can't $1^4+4\cdot1^4$ be a prime number? – George Law Aug 28 '16 at 21:31
• Maybe it should be $p>5$. – iamvegan Aug 28 '16 at 21:33

$$a^4 + 4 b^4 = \left( {a}^{2}-2\,ab+2\,{b}^{2} \right) \left( {a}^{2}+2\,ab+2\,{b}^ {2} \right)$$ so, with one notable exception, this can't be prime.
Start from the following $$p = a^4+4b^4 = (a^2+2b^2)^2-4a^2b^2 = (a^2+2ab+2b^2) (a^2-2ab+2b^2).$$ So $p$ can be prime only if $$a^2-2ab+2b^2=1.$$ Which corresponds to $a=b=1$ and $p=5$.