<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] I think that hint is terrible and only the geniuses in the class could get it right away. I put $(x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$ into Wolfram Alpha and it gave me $x^4 + 4y^4$ as an "alternate form."
But the problem says "$4^b$", not "$4y^4$." If you're a precocious genius you can right away see that if $b$ is odd then $$4^b = 4 \left(2^{\frac{2b - 2}{4}} \right)^4.$$ I'm not a genius, so it took me a while (three days, to be precise) to arrive at this. It was only then that the problem finally unraveled. Now we have $$a^4 + 4 \left(2^{\frac{2b - 2}{4}} \right)^4 = (a^2 - 2^{\frac{2b - 2}{2}} a + 2^b)(a^2 + 2^{\frac{2b - 2}{2}} a + 2^b).$$
This means that with $b$ odd, the only way for the resulting number to be prime is for $(a^2 - 2^{\frac{2b - 2}{2}} a + 2^b)$ to equal $1$ or $-1$ somehow. There are four possible solutions to that equation, and they're all ruled out with the definition of the problem (because $a$ and $b$ are natural numbers not both equal to $1$, and $0$ is or isn't a "natural number" -- but that's a whole other can of worms). Therefore, under the constraints of the problem, odd $b$ gives a composite number, so $b$ must be even in order for the expression to give a prime (this is a necessary but not sufficient condition, of course).
Let's have a concrete example: $a = 1$, $b = 3$. Then $a^4 + 4^b = 65$, which is obviously composite. But still take the time to run it through Sophie Germain's identity: $1 + 4^3 = 1 + 4 (2^1)^4 = (1 - 2 \times 1 \times 2 + 2 \times 2^2)(1 + 2 \times 1 \times 2 + 2 \times 2^2) = 5 \times 13$.
As for the parity of $a$, that, as Bob Happ said in a comment, "is the easy part here."
</s>