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I am unable to proceed with the following problem, : If $a+b+c=0$ then find all the ordered pairs $(m,n)$ where $m,n\in\mathbb{N}$ such that $$\boxed{\frac{a^{m+n}+b^{m+n}+c^{m+n}}{m+n}=\frac{(a^m+b^m+c^m)}{m}\frac{(a^n+b^n+c^n)}{n}}$$

I tried but I think I have no idea regarding this kind of problem although I tried a bit hard with all possible methods (which I can do/know) but that doesn't result any substantial thing which I can add and also it will not give a complete set of solutions, this trial approach provides a few solution(and quite laborious and too many limitations) for instance this well known stuff if $a+b+c=0$ then, $\frac{a^5+b^5+c^5}{5}=(\frac{a^3+b^3+c^3}{3})(\frac{a^2+b^2+c^2}{2})$ is one of the solution. Thanks for your attention.

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    $\begingroup$ Inquisitive you are, be a little attemptive as well! Can you think of obvious pairs which work? Which don't work? Take examples : for example take $a = -b$ and $c=0$ to simplify things, then what works? $\endgroup$ Commented Jan 4, 2021 at 6:15
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    $\begingroup$ @teresa but it will not give a complete set of solutions, this trial approach provides a few solution(and quite laborious and too many limitations) for instance this well known stuff if $a+b+c=0$ then $\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\frac{a^2+b^2+c^2}{2}$ For your concern! yeah, I have done that but it was not quite good anyway $\cdots$ $\endgroup$
    – random
    Commented Jan 4, 2021 at 9:41
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    $\begingroup$ What is your source for this problem? Also look up symmetric polynomials, the Newton-Girard formula as well, they seem related. $\endgroup$ Commented Jan 4, 2021 at 9:43
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    $\begingroup$ @ teresa but I can't proceed and connect the stuffs, can you help a bit more? $\endgroup$
    – random
    Commented Jan 6, 2021 at 8:40
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    $\begingroup$ That is fine, I want to know the source for your problem. $\endgroup$ Commented Jan 6, 2021 at 8:48

1 Answer 1

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Let us first consider the case $m\le n$.

  • If $m=n$, then taking $(a,b,c)=(1,-1,0)$, we get $m=(1+(-1)^m)^2$ from which $m=4$ follows. However, when $m=n=4$, the equation does not hold for $(a,b,c)=(2,-1,-1)$.

  • If $m=1$, then taking $(a,b,c)=(2,-1,-1)$, we get $2^{n}=(-1)^{n}$ which is impossible.

So, in the following, $2\le m\lt n$.

  • If both $m$ and $n$ are odd, then taking $(a,b,c)=(2,-1,-1)$, we get $$mn(2^{m+n-1}+1)=2(m+n)(2^{m-1}+(-1)^m)(2^{n-1}+(-1)^n)$$The LHS is odd while the RHS is even, which is impossible.

  • If both $m$ and $n$ are even, then taking $(a,b,c)=(1,-1,0)$, we get $(m-2)(n-2)=4$, but there are no solutions satisfying $2\le m\lt n$.

  • If exactly one of $m,n$ is odd, then taking $(a,b,c)=(2,-1,-1)$, we get $$mn(2^{m+n}-2)=(m+n)(2^m+2(-1)^m)(2^n+2(-1)^n)\tag1$$Suppose here that $m$ is odd. Then, since we have $mn\gt m+n\ (\gt 0)$ (which is equivalent to $(m-1)(n-1)\gt 1$ which is true) and $$2^{m+n}-2\gt (2^m+2(-1)^m)(2^n+2(-1)^n)\ \ (\gt 0)$$ (which is equivalent to $2\gt 2^{m+1}-2^{n+1}$ which is true), the LHS of $(1)$ is larger than the RHS of $(1)$. So, $m$ has to be even to have $$2^n\bigg(\underbrace{n(m2^{m}-2^{m}-2)-2^mm-2m}_{A}\bigg)+\underbrace{n(2^{m+1}-2m+4)+4m(2^{m-1}+1)}_{\gt 0}=0\tag2$$Now, suppose that $m\ge 4$. Then, we have $$\begin{align}A=n\underbrace{(m2^{m}-2^{m}-2)}_{\gt 0}-2^mm-2m&\gt m(m2^{m}-2^{m}-2)-2^mm-2m \\&=4m\bigg((m-2)2^{m-2}-1\bigg) \\\\&\gt 0\end{align}$$implying that the LHS of $(2)$ is positive. So, we have to have $m=2$. Taking $(a,b,c)=(2,-1,-1)$, we have$$(n-6)2^{n-1}+2n+6=0\tag3$$Suppose here that $n\ge 7$. Then, the LHS of $(3)$ is positive. So, we get $n=3,5$ which are sufficient (see here, here, here).

In conclusion, considering the case $m\gt n$, we see that the answer is $$\color{red}{(m,n)=(2,3),(3,2),(2,5),(5,2)}$$

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  • $\begingroup$ Excellent answer !! (+1) $\endgroup$
    – Spectre
    Commented Feb 1, 2021 at 6:02

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