This question was inspired by the post here, which asks for a proof of the following fact:

If $a+b+c = 0$ then show that $$\frac{a^3 + b^3 + c^3}{3} \cdot \frac{a^7 + b^7 + c^7}{7} = \left( \frac{a^5 + b^5 + c^5}{5} \right)^2$$

I was curious about finding all exponents for which this works, arriving at the following problem:

For which distinct positive integers $m,n$ (with $m+n$ even) is it true that

$$\frac{a^m + b^m + c^m}{m} \cdot \frac{a^n + b^n + c^n}{n} = \left( \frac{a^\frac{m+n}{2} + b^\frac{m+n}{2} + c^\frac{m+n}{2}}{(m+n)/2} \right)^2$$

for all $a,b,c$ with $a+b+c = 0$?

I added the distinctness condition because $m=n$ always trivially works.

So far, the only solution I know of is $(3,7)$, as per the post quoted. A Mathematica search yielded that there are no other solutions with $a,b \leq 100$.

A solution for this problem would of course be really nice, but I don't really have high hopes for one! The shortest proof of the $(3,7)$ case uses Newton's formulas for power sums, which are defined recursively and don't have a nice closed form, as far as I know. So any kind of intuition/heuristic about why there should/shouldn't be any more solutions would also be very interesting.

Some progress: I decided to run a much bare-bone simulation in Mathematica, by checking if the identity holds at least for $(a,b,c) = (1,-\frac{1}{2},-\frac{1}{2})$, and indeed for $m,n \leq 2000$ it only holds for $(m,n) = (3,7)$. Plugging in, we have

$$\frac{1}{mn} \left(1 + \frac{(-1)^m}{2^{m-1}}\right)\left(1 + \frac{(-1)^n}{2^{n-1}}\right) = \frac{4}{(m+n)^2}\left( 1 + \frac{(-1)^\frac{m+n}{2}}{2^{\frac{m+n}{2} - 1}} \right)^2.$$

Thus we can turn the problem into a Diophantine equation, which might make things simpler?

  • $\begingroup$ Meant to say $m+n$ cannot be odd, so they have to be both odd or even, otherwise, you're forced to take the square root of a negative number since $a+b+c=0$ $\endgroup$ Apr 13, 2021 at 23:19
  • $\begingroup$ Indeed, I edited the question to reflect that. $\endgroup$ Apr 13, 2021 at 23:22
  • $\begingroup$ Have you tried using the idea of the S.C.B. solution from the link in this problem? $\endgroup$
    – jjagmath
    Apr 14, 2021 at 1:05

1 Answer 1


The first step is to pin down the parity of $m$ and $n$. Since $m+n$ is even, we must have $m\equiv n\pmod{2}$.

Considering the case $(a,b,c)=(x,-x,0)$, we have $$x^{m+n}\cdot\frac{(1+(-1)^m)(1+(-1)^n)}{mn}=x^{m+n}\cdot\frac{4\left(1+(-1)^{\frac{m+n}{2}}\right)^2}{(m+n)^2}$$ Canceling $x$ and considering when these vanish, we find $m\equiv n\equiv\frac{m+n}{2}\pmod{2}$.

In the even case, we simplify to obtain $$\frac{1}{mn}=\frac{4}{(m+n)^2}\text{;}$$ thus $4mn=(m+n)^2$. Rearranging, $(m-n)^2=0$. (The original version of this paragraph had an error; HT to Tanny Sieben for pointing out the appropriate correction.)

In the odd case, things are more tricky. Following your idea, let's take $(a,b,c)=(2,-1,-1)$. (Sorry; fractions are annoying to type.)

Then $$\frac{(2^m-2)(2^n-2)}{mn}=\left(\frac{2^{\frac{m+n}{2}}-2}{\frac{m+n}{2}}\right)^2$$ Again expanding, we have $$(m+n)^2(2^{m+n}-2^{n+1}-2^{m+1}+4)=4mn(2^{m+n}-2^{\frac{m+n}{2}+2}+4)$$ Rearranging via the identity $(m+n)^2-4mn=(m-n)^2$, we have $$(m-n)^2(2^{m+n}+4)=(m+n)^2(2^{n+1}+2^{m+1})-mn2^{\frac{m+n}{2}+4}$$

At this point, a change of variables suggests itself: let $a=\frac{m+n}{2}$ and $d=\frac{m-n}{2}$. Note that $a>d>0$, $a$ is odd, and $d$ is even. In any case, $$4d^2(2^{2a}+4)=4a^2(2^{a+d+1}+2^{a-d+1})-(a^2-d^2)2^{a+4}$$ Canceling a factor of $2^{a+3}$, $$d^2(2^{a-1}+2^{1-a})=a^2(2^d+2^{-d})-2(a^2-d^2)$$ Rearranging, $$d^2(2^{a-1}+2^{1-a}-2)=a^2(2^d+2^{-d}-2)$$

But our equation is now a perfect square: $$d^2\left(2^{\frac{a-1}{2}}-2^{\frac{1-a}{2}}\right)^2=a^2\left(2^{\frac{d}{2}}-2^{-\frac{d}{2}}\right)^2$$ Taking square roots, and rearranging, $$\frac{2^{\frac{a-1}{2}}-2^{\frac{1-a}{2}}}{a}=\frac{2^{\frac{d}{2}}-2^{-\frac{d}{2}}}{d}\tag{1}$$

Call these $L(a)$ and $R(d)$, and consider them as functions over $\mathbb{R}^+$. What can we say?

Well, both $L$ and $R$ are increasing to infinity, with $\lim_{a\to0^+}{L(a)}=-\infty$ and $\lim_{d\to0^+}{R(d)}=\ln{(2)}$. Thus, for each $d$, there is a unique $a$ satisfying (1). Checking some small integers, we find (as expected) that there is no solution for $d\in\{1,3,4,5\}$, but there is, of course, $(d,a)=(2,5)$.

Moreover, for $d\geq5$, we have $L(d+2)>R(d)$. Thus $d<a<d+2$; that is to say, if there is another integer solution, it arises from $a=d+1$.

Are there any such solutions? Well, if $a=d+1$, then $n=1$ (w/oLoG) and $m=2d+1$. So $$\left(\frac{a^{d+1}+b^{d+1}+c^{d+1}}{d+1}\right)^2=\left(\frac{a^{2d+1}+b^{2d+1}+c^{2d+1}}{2d+1}\right)\left(\frac{a+b+c}{1}\right)$$ But the latter is $0$ by assumption (!). It is easy to see then that we must have $d=0$, the trivial case.

  • 1
    $\begingroup$ This is fantastic! I think that in the even case, when cancelling $x^{m+n}$ the resulting equation should be $\frac{1}{mn} = \frac{4}{(m+n)^2}$ (there's an extra factor of 4 on the LHS), which reduces to $(m+n)^2 = 4mn \Rightarrow (m-n)^2 = 0$, giving the trivial case. The rest of the solution is really nice, thanks! $\endgroup$ Apr 14, 2021 at 10:08

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