A couple days ago, someone posted a question about using integer solution to the equation $a^a+b^b=c^c$ to disprove Fermat's last theorem. The question has since been deleted but I was curious as to whether or not there are integer solutions to the equation.

More importantly though, the mention of Fermat's last theorem got me thinking. Might there be an analogue of Fermat's last theorem for tetration?

Specifically, it occurs to me the equation $a^a+b^b=c^c$ is the case of ${^n}a+{^n}b={^n}c$ for which $n=2$. If there are integer solutions for $n=2$, then what about $n=3$? $n=4$? $n=1397$?

I would guess that there are only integer solutions if $n=1$, but is this the case? And what about subsequent hyperoperations?

  • $\begingroup$ Since it is known that in fermat's last theorem there are no solutions for $n>2$ Them maybe $a,b,c<2$ would be the only solution to this? Although the two are not directly related as all 3 terms are not raised to the same power $\endgroup$ – Henry Lee Apr 22 '19 at 15:16
  • $\begingroup$ @PeterForeman I've seen people do some pretty amazing things with large numbers and modular arithmetic. Number theory isn't my strong suit, though so I'm definitely not the one to ask. $\endgroup$ – R. Burton Apr 22 '19 at 15:22
  • $\begingroup$ Whyt do you mean by $^{n}a$? $\endgroup$ – amsmath Apr 22 '19 at 15:25
  • $\begingroup$ @amsmath Tetration: en.wikipedia.org/wiki/Tetration $\endgroup$ – R. Burton Apr 22 '19 at 15:26
  • $\begingroup$ @R.Burton I see. Thank you. $\endgroup$ – amsmath Apr 22 '19 at 15:30

WLOG assume $a\leq b$ then $c\geq b+1$ and $^nc\geq {^n} (b+1)>{^n} b+{^n} b\geq{^n} b+{^n} a$.

  • $\begingroup$ What if $a<0$ or $b<0$? $\endgroup$ – R. Burton Apr 22 '19 at 15:31
  • $\begingroup$ If a, b are even it doesn't matter otherwise if $b<0$ then $c<0$ then you have ${^n} a+{^n} (-c) =^{n} (-b) $. If $a<0$ then ${^n} b={^n} c+{^n} (-a) $. $\endgroup$ – kingW3 Apr 22 '19 at 15:42
  • $\begingroup$ ... *for $n \gt 1$ " ... $\endgroup$ – Gottfried Helms May 16 '19 at 0:37

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