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.
 A: This boils down to showing that $b^4 + b + 1$ is never the fourth power of an integer.
The difference between consecutive fourth powers is
$$(b+1)^4 - b^4 = 4b^3 + 6b^2 + 4b + 1,$$
which is greater than $b+1$ for all $b > 0$.
Since $b^4 + b + 1$ is a fourth power of an integer, plus $b+1$, you'll never reach another fourth power of an integer by adding $b+1$, because you'll never reach the next fourth power of an integer.
A: Indeed, we can show that $b^4+b+1=c^2$ has no solution in positive integers.
This is because if $b>0$ is an integer, then $0<b+1<2b+1\leq 2b^2+1$.  Adding $b^4$ to this inequality gives: so $$b^4<c^2=b^4+b+1<b^4+2b^2+1=(b^2+1)^2$$ so $$b^2<c<b^2+1.$$
There are also no solutions with $b\leq -2$, since then:
$$1-2b^2<b+1<0$$ and then add $b^4$ to get: $$(b^2-1)^2<b^4+b+1<b^4$$
There is a $c$ with $b=0,-1$.
A: 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!
