Prove that there are no positive integers $a$ and $b$, such that $b^4 + b + 1 = a^4$ So I've been trying to solve this problem for a couple of days now. What I've come up with is this:
By way of contradiction, assume that there are positive integers a and b such that $b^4 + b + 1 = a^4$.
Consider the case when $b \geq a$. Then $a^4 - b^4 = b + 1$, but also $a^4 - b^4 \leq 0$. Thus, $b + 1 \leq 0$. Then, $b \leq -1$, a contradiction.
However, I haven't been able to figure out how to prove it with the case of $b < a$.
Thank you.

In any case, $b^4-a^4+b+1=0$. Also,
$$
(b^4-a^4)=(b^2-a^2)(b^2+a^2)=(b-a)(b+a)(b^2+a^2)
$$
So, 
$$
(b-a)(b+a)(b^2+a^2)= -b-1
$$
So $(b^2+a^2)$ now divides $b+1$. So $b^2+a^2\leq b+1$. The only way this will work is if $a=b=1$. Check that this doesn't satisfy the original equation. So we're done!
