<s>[INST] 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.
 [/INST] 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 $b=0,-1$.
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