Diophantine power equation of degree 4 and 5 and 2 variables

Prove that $a^4 + 1 = 2b^4$ and $a^4 - 1 = 2b^4$ have no solutions in integers. Same with $a^5$ and $b^5$.

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Are $0$ and $1$ no longer considered integers? –  Erick Wong Jan 7 at 22:33
@user55514: Welcome to MSE. Generally, you should only ask one question at a time. Also, it would help if you would formally write it out for your second question as to not have people guessing. Additionally is this a homework problem? If so, please tag it as such. Regards –  Amzoti Jan 7 at 22:33
You might get better answers if you were to show what you have tried, or at least give some background about where this question originated (that helps to determine which tools might be available to answer the question). –  robjohn Jan 8 at 0:51
Perhaps you meant: "Prove that neither $a^4 + 1 = 2b^4$ nor $a^4 - 1 = 2b^4$ have solutions in integers greater than $1$." –  robjohn Jan 8 at 0:55
For the first one, see my answer here: mathoverflow.net/questions/24609/… –  Byron Schmuland Jan 8 at 14:49

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 $$[1] \ \ t(t+1)(2t^2+2t+1)=4s^4.$$ 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 [1] 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.

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Thanks coffeemath. Hope somebody will find a not too hard way to solve a^4 +1 = 2b^4 –  user55514 Jan 8 at 14:41
Thank you Byron. It would be of some interest if considering now the general equation a^2n +1 = 2b^2n; if the method could be extended to prove no solutions to z^2 = ((a^2n -1)/2)^2 = b^4n -a^2n. –  user55514 Jan 8 at 17:45
$$a^4+1=2b^4\,\,,\,\,a^4-1=2b^4\Longrightarrow a^4+1=a^4-1$$
and the last equation has no solution in any field (ring) of characteristic different from $\,2\,$ , let alone in the integers.
Although the grammar may not be correct, I don't think that the intended question was "Prove that $a^4+1=2b^4$ and $a^4-1=2b^4$ do not have simultaneous solutions". I think it is clear that the OP intended "Prove that neither $a^4 + 1 = 2b^4$ nor $a^4 - 1 = 2b^4$ have solutions in integers." Just as people whose native language is not English struggle to post questions in English, we should make some effort to understand their questions, even if their translation is not quite right. –  robjohn Jan 8 at 0:28