# Ordered pairs $(m,n)$ such that ${\frac{a^{m+n}+b^{m+n}+c^{m+n}}{m+n}=\frac{(a^m+b^m+c^m)}{m}\frac{(a^n+b^n+c^n)}{n}}$.

I 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.

• 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? Commented Jan 4, 2021 at 6:15
• @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 $\cdots$ Commented Jan 4, 2021 at 9:41
• What is your source for this problem? Also look up symmetric polynomials, the Newton-Girard formula as well, they seem related. Commented Jan 4, 2021 at 9:43
• @ teresa but I can't proceed and connect the stuffs, can you help a bit more? Commented Jan 6, 2021 at 8:40
• That is fine, I want to know the source for your problem. Commented Jan 6, 2021 at 8:48

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 $$\color{red}{(m,n)=(2,3),(3,2),(2,5),(5,2)}$$

• Excellent answer !! (+1) Commented Feb 1, 2021 at 6:02