(Sorry if the title isn't clear. I couldn't think of a good title for this question.)

Edit: @mathlove pointed out some mistakes in my original question. We can easily see anything that are in the criterion of Fermat's last theorem can be disproved (for m=n$\geq$3.) what about for those which doesn't satisfy the criteria in Fermat? e.g. $a^{128}+b^2$?

This is a question concerning the properties of exponentiation. Say a certain constant, $C$ is a part of a diophantine equation, e.g. $a^2+b^2=C$ that (we are sure) has integer roots $a$, $b$. We then take the constant $C$ to the exponent of itself, $C^C$. Will the new value, $C^C$ have the an integer solution to the same equation? (while the actual integer root values might change, I'm simply concerned with thether I can have an integer solution).

The question can be shortened into this form: given $C$ and a random diophantine equation of two variables, which, on the right hand side has $C$ as a constant and on the left hand side the general form of $a^n+b^m$. Assume we know there exists integer solutions $a$, $b$ to the diophantine equation, can we conclude that an integer solution also exists for the same diophantine equation with $C$ replaced as $C^C$?

  • $\begingroup$ This is equal to asking whether if $(a^m+b^n)^{(a^m+b^n)}$ can be written in terms of $(p^j+q^k)$, where $p,q,j,k$ are integers. $\endgroup$ – BearAqua Mar 28 '17 at 14:52
  • $\begingroup$ $a^4+b^4=2$ holds for $a=b=1$, but there are no integers $a,b$ such that $a^4+b^4=2^2$. And note that $13^{13}\not=169$. $\endgroup$ – mathlove Mar 28 '17 at 15:10
  • $\begingroup$ No, there is no immediate reason for this to hold and I guess that it only holds exceptionally. (Equations with $n=1$ or $m=1$ are ruled out as they always have integer solutions.) $\endgroup$ – Yves Daoust Mar 28 '17 at 15:26
  • $\begingroup$ @mathlove Thanks for pointing out. I 've revised it a bit. I guess I'm really concerned with cases that do not satisfy $m=n\geq3$. $\endgroup$ – BearAqua Mar 28 '17 at 15:31
  • $\begingroup$ And, additionally, for my comment on top, it should satisfy $j=m$ and $k=n$ $\endgroup$ – BearAqua Mar 28 '17 at 15:37

The short answer is no (see the comments above). Here are some counterexamples collected from the comments thread (with thanks to @mathlove for pointing them out):

  1. $𝑎^4+𝑏^4=2$ holds for $𝑎=𝑏=1$, but there are no integers $𝑎,𝑏$ such that $𝑎^4+𝑏^4=22$.

  2. $𝑎^{128}+𝑏^2=5$ holds for $𝑎=1,𝑏=2$, but there are no integers $𝑎,𝑏$ such that $𝑎^{128}+𝑏^2=55$.


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