Under what all conditions on $(a,b,c)$ where $a,b,c$ are positive integers can we say about the non-existence or existence of any solution to the equation $$a^a+b^b=c^c$$

(Or in other words : Solve $a^a+b^b=c^c$ ) $ $

$ $

About this note that : For a solution we must have $gcd(a,b,c) = 1$ (follows from Fermat's Last Theorem)

and all the elements of the set $\{a,b,c\}$ cannot be prime numbers (proof shown below).

We will prove there are no positive prime numbers $p,q,r$ such that the following equation is satisfied : $$p^p+q^q=r^r$$ This is easy to prove. Note that in fact we have something stronger : The equation $4+m^m=n^n$ has no solutions in positive integers. Assume the contrary, then we get $m>1$ (by checking the case $m=1$ separately) and then note that $n \ge m+2$ and thus $n^n > (m+2)^m > m^m+2^m \ge n^n$ and thus a contradiction.

  • $\begingroup$ I wondered about what other variants of your equation can be interesting, from the answer's remark about how grows the sequence $n^n$. As we know $f(n)=n$ in your exponent and basis is a multiplicative and additive function. I've considered the (I refer Wikipedia) Prime omega function $\omega(n)$ that counts the number of distinct prime factors in the factorization of $n>1$. My proposal was $a^{\omega(a)}+b^{\omega(b)}=c^{\omega(c)}$ with $a<b<c$ (and now I believe that also I need to require $1<a$ to get a problem more interesting problem). I know also the MSE post 1861380 $\endgroup$
    – user142929
    Jan 31, 2020 at 12:41

1 Answer 1


The sequence $(n^n)_{n \ge 1}$ grows too fast for there to exist a solution: if $b \ge 2$, then $(b+1)^{b+1} > b^{b+1} \ge 2 b^b$ (and also $2^2 > 2 \cdot 1^1$), so if $1 \le a \le b$, then $b^b < a^a + b^b \le 2 b^b < (b+1)^{b+1}$, and no $c$ can exist such that $a^a + b^b = c^c$.

  • 4
    $\begingroup$ Didn't even need Fermat's Last Theorem! $\endgroup$ May 7, 2017 at 15:08
  • 1
    $\begingroup$ Excellent. I can see this as a great trick problem to put on a number theory exam. $\endgroup$
    – Thompson
    May 7, 2017 at 17:55
  • 2
    $\begingroup$ On the other hand, the equation $x^xy^y=z^z$ is more interesting. $\endgroup$
    – Pasten
    May 8, 2017 at 1:17
  • $\begingroup$ Ok yes friends and teachers and everyone here. I am obliged that you found it interesting. Anyways, I thought it like a difficult problem and hence used the name FLT (which I never read the proof because I don't know enough reqd. for knowing it) so yes appreciate @GeoffreyIrving. $\endgroup$ May 9, 2017 at 11:22
  • 1
    $\begingroup$ Just to clarify: the equation $x^xy^y=z^z$ is a problem of Erdös. See problem D13 of R. Guy's book "Unsolved problems in number theory, 2nd edition" for a reference $\endgroup$
    – Pasten
    May 9, 2017 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.