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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 $a^a+b^b=c^c$ ) $ $

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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.

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  • $\begingroup$ 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

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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 $a^a + b^b = c^c$.

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    $\begingroup$ Didn't even need Fermat's Last Theorem! $\endgroup$ May 7, 2017 at 15:08
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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