<s>[INST] If $a^4 + 4^b$ is prime, then $a$ is odd and $b$ is even. We say an integer $p>1$ is prime when its only positive divisors are $1$ and $p$. Let $a$ and $b$ be natural number not both $1$. Prove that if $a^4+4^b$ is prime, then $a$ is odd and $b$ is even.
I'm also given a hint:
Consider the expression $(x^2-2xy+2y^2)(x^2+2xy+2y^2)$.
What I have so far is:
Let $x=a^2, y=2^b$ and substitute it into the expression.
But I am not sure how to proceed. Can anyone help please?
 [/INST] It's clear that $a$ must be odd, as otherwise the number would be divisible by $2$.
Suppose $b$ is odd. Then we can write our number as $a^4 + 4\cdot 4^{2n} = a^4 + 4 \cdot 2^{4n}$ for some $n$. You were given the key factorization,
$$ (a^4 + 4 b^4) = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2).$$
here, that means
$$ a^4 + 4 \cdot 2^{4n} = (a^2 + 2 a 2^n + 2\cdot  2^{2n})(a^2 - 2 a 2^n + 2\cdot  2^{2n}),$$
which is a nontrivial factorization. Thus $b$ cannot be odd.
Aside:
This factorization identity is called Sophie Germain's identity, after the French mathematian and general polymath who noticed it while exploring number theory. She corresponded with Gauss and Legendre, and attended lectures at the newly founded École Polytechnique. To avoid ridicule, as it was extremely uncommon for women to pursue math (or indeed, to be allowed to pursue a great many professions), she used a male pen name for her initial correspondence.
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