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 '13 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 '13 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 '13 at 0:55
For the first one, see my answer here: mathoverflow.net/questions/24609/… – Byron Schmuland Jan 8 '13 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 '13 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 '13 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.

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I think it should be proved that neither the first nor the second equation has a solution in integers except (1,0) – miracle173 Jan 8 '13 at 0:07
As written, it looks like a system of two equations, which as proved has no solution. If the OP wanted the equations separatedly he could have used "neither" or "each" or something like that, imo. – DonAntonio Jan 8 '13 at 0:16
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 '13 at 0:28
Well, I'm not a native english speaker and, in fact, english is just my third language (it shows, uh?), so I may have not understood the actual intention of the OP...Yet he hasn't still said a word about anything after posting his question. – DonAntonio Jan 8 '13 at 0:37
It's only been a couple of hours. Perhaps the OP did not expect such a quick response. There are probably good reasons that one might not be able to attend to their account for a few hours. I like to give the benefit of the doubt, especially to new users. – robjohn Jan 8 '13 at 0:46