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Can $\frac{a^n+b^n}{a+b}=c^n$ have a solution for $a,b,c,n\in \mathbb N$, $a^n+b^n\neq a+b$, and $\gcd(a+b,c)=\gcd(a,b)=1$?

I notice that $n$ must be odd for the LHS to be integer. Also, as $\frac{a^n+b^n}{a+b}<(a+b)^{n-1}$, it follows that $c<a+b$; but little more progress...

Thanks in advance for your hints and advice!

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    $\begingroup$ "$n$ must be odd for the LHS to be integer". Take $$ \frac{1^2 +1^2}{1+1} = 1^2$$ and it indeed holds that $n>1$ and $\gcd(2,1)=1$. $\endgroup$
    – Robert Lee
    Commented Sep 25, 2021 at 21:48
  • $\begingroup$ @RobertLee well spotted! However, I am looking for non-trivial solutions, where $a^n+b^n\neq a+b$. I thought that with the constraint $n>1$ that was already settled, so I will edit the OP if you don't mind $\endgroup$ Commented Sep 26, 2021 at 5:54
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    $\begingroup$ Certainally when $n=3$ there are other solutions. For example each of the following triples $(a,b,c)$ work $ (0, -1, 1), (0, 1, 1), (-1, -1, 1), (1, 1, 1), (-1, 0, 1), (1, 0, 1), (18, -1, 7), (-1, 18, 7), (-18, 1, 7), (1, -18, 7), (-19, -18, 7), (-19, -1, 7), (18, 19, 7), (19, 1, 7), (1, 19, 7), (-1, -19, 7), (-18, -19, 7), (19, 18, 7), (-17, 36, 13), (17, -36, 13), (36, -17, 13), (-36, 17, 13), (17, 53, 13), (-17, -53, 13), (36, 53, 13), (-36, -53, 13), (-53, -36, 13), (53, 17, 13), (53, 36, 13), (-53, -17, 13)$ I'd definitely guess there are infinitely many when $n=3$. $\endgroup$ Commented Sep 26, 2021 at 11:18
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    $\begingroup$ @Juan well I only thought about the case when $n$ is odd. In that case the LHS is a polynomial of degree $n-1$. This equation then cuts out a curve in $\mathbb{P}(n,n,n-1)$ (weighted projective space). Solutions should conincide(ish) with rational points on this curve (call it $C_n$). When $n \geq 5$ this curve has genus $\geq 2$, so has only finitely many rational points by Faltings' Theorem. In practice when one searches for points on curves of genus $\geq 2$ with nice enough equations you usually find all the solutions straight away (at least if you use clever enough point searching) $\endgroup$ Commented Sep 26, 2021 at 18:34
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    $\begingroup$ When $n=3,$ there are infinitely many positive integer solutions as follows. $(a,b,c)=(p^3-3pq^2+q^3, 3p^2q-3pq^2, p^2-pq+q^2)$ with $p,q$ are arbitrary integers. $\endgroup$
    – Tomita
    Commented Sep 27, 2021 at 0:22

1 Answer 1

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For, $n=2$, we have:

$a^2+b^2=(a+b)c^2$ -------$(1)$

Above has numerical solution:

$(a,b,c)=(4,28,5)$

Also in eqn(1) if we take:

$a^2+b^2=c^2$, then the condition needed is:

a+b=1

We take:

$(a,b,c)=[(2mn),(m^2-n^2),(m^2+n^2)]$

Hence the condition is:

$m^2+2mn-n^2=1$

Above has solution at, $(m,n)=(1,2)$

And we get:

$(a,b,c)=[(4),(-3),(5)]$

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