Prove that $a^4 + 1 = 2b^4$ and $a^4 - 1 = 2b^4$ have no solutions in integers. Same with $a^5$ and $b^5$.
This addresses the equation $a^4-1=2b^4$. Note we may assume $a,b \ge 0$ since changing their signs has no effect, and sign possibilities may be noted afterwards. Now note that $a$ must be odd and write $a=2t+1$ where $t \ge 0$ is an integer. Then the equation in terms of $t$, on factoring $a^4-1$ and dividing by 2, becomes $$4t(t+1)(2t^2+2t+1)=b^4.$$
Now we see $b$ is even and so we can put $b=2s$, so $b^4=16s^4$, and we arrive at $$ \ \ t(t+1)(2t^2+2t+1)=4s^4.$$ The three factors on the left are pairwise coprime. That $\gcd(t,t+1)=1$ and $\gcd(t,2t^2+2t+1)=1$ is immediate, and if a prime $p$ were to divide both $t+1$ and $2t^2+2t+1$, then $p$ would also divide $(t+1)^2=t^2+2t+1$, and hence also divide the difference $(2t^2+2t+1)-(t^2+2t+1)=t^2.$ but then $p$ divides both $t+1$ and $t$, impossible.
So equation  is a product of three pairwise coprime factors equal to the square $(2s^2)^2$. Therefore all three factors are squares. But the only way both $t$ and $t+1$ are square is if $t=0$, leading to the solution $(a,b)=(1,0)$, where we can also include $(a,b)=(-1,0)$ on changing sign. These are then the only integer solutions to $a^4-1=2b^4$. The same argument works for the more inclusive equation $a^4-1=2b^2$, by the way. The other equation $a^4+1=2b^4$ has not such a simple solution, at least that I can see.
and the last equation has no solution in any field (ring) of characteristic different from $\,2\,$ , let alone in the integers.
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