For $a, b, c$ with $a+b+c=0$ prove $\frac15\sum a^5=\frac13\sum a^3\cdot\frac12\sum a^2$ and $\frac17\sum a^7=\frac15\sum a^5\cdot\frac12\sum a^2$

Question

Consider the following problem:

Problem. Suppose that real numbers $a$, $b$ and $c$ satisfy the condition $a+b+c=0$. Prove the following identities: $$ \frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot\frac{a^2+b^2+c^2}{2}, \\ \frac{a^7+b^7+c^7}{7}=\frac{a^5+b^5+c^5}{5}\cdot\frac{a^2+b^2+c^2}{2}. $$

Perhaps, the shortest solution I can think of is as follows: plug $c=-a-b$ into the equation but instead of expanding everything use the following identities $$ (a+b)^3-a^3-b^3=3ab(a+b), \\ (a+b)^5-a^5-b^5=5ab(a+b)(a^2+ab+b^2), \\ (a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2. $$ However, these identies are coming out of nowhere and moreover, in order to prove them one still needs to do some computations.

Question. Is it possible to solve this problem in a "smart" way (i.e. avoiding computations and preferrably elementary since it is almost a high school problem)? Any other solutions are also welcome.

Comment. It should be also noted that it is unclear (at least for me) how those identies were invented. It seems there is no nice similar identities for $\frac{a^p+b^p+c^p}{p}$ for $p$ other than 2, 3, 5, 7.

Answer 1

This is a problem about elementary symmetric functions. Let $$s_1=a=b+c$$ $$s_2=ab+bc+ac$$ $$s_3=abc$$

Now assuming that $s_1=a+b+c=0$ we show that

\begin{equation*} \begin{aligned} -\frac{a^2+b^2+c^2}{2}&=s_2=ab+bc+ca\\ \frac{a^3+b^3+c^3}{3}&=s_3=abc\\ -\frac{a^4+b^4+c^4}{4}&=-\frac{1}{2}s_2^2=-\frac{(a^2+b^2+c^2)^2}{2}\\ \frac{a^5+b^5+c^5}{5}&=-s_2s_3\\ -\frac{a^6+b^6+c^6}{6}&=-\frac{1}{2}s_3^2+\frac{1}{3}s_2^3\\ -\frac{a^7+b^7+c^7}{7}&=s_2^2s_3\\ \end{aligned} \end{equation*}

And the results follow.

The calculational scheme for deriving the above equalities is

to note that

$$\ln (1+ax)=ax-\frac{a^2}{2}x^2+\frac{a^3}{3}x^3+\cdots $$ and adding,

$$\ln (1+ax)(1+bx)(1+cx)= (a+b+c)x-\frac{a^2+b^2+c^2}{2}x^2+ \frac{a^3+b^3+c^3}{3}x^3+\cdots $$

on the other hand,

$$(1+ax)(1+bx)(1+cx)= 1+s_1x+s_2x^2+s_3x^3$$ so we have under the assumption that $s_1=0$,

$$\ln (1+s_2x^2+s_3x^3)=(s_2x^2+s_3x^3)-\frac{1}{2} (s_2x^2+s_3x^3)^2+\frac{1}{3}(s_2x^2+s_3x^3)^3\cdots $$ from which the above equalities follow easily by equating the coefficients of the two power series.

In particular this shows that every $\frac{a^n+b^n+c^n}{n}$ is a polynomial in $n=2,3$ assuming $s_1=0$.

Answer 2

let's have a look at the following identities called Lame-type identities (see here).

$$(x+y+z)^3 - (x^3+y^3+z^3) = 3(x+y)(x+z)(y+z)$$

$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)$$

$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))$$

Algebraically manipulating the second and third identities give;

Let $$A = x+y+z,$$ $$B = x^2+y^2+z^2,$$ $$C = x^3+y^3+z^3,$$ $$E = x^5+y^5+z^5,$$ $$G = x^7+y^7+z^7,$$ then
\begin{equation} 6E=A^5-5BA^3+5CA^2+5BC \tag{1} \end{equation}

\begin{equation} 36G=A^7+7CA^4-21B^2A^3+28C^2A+21B^2C \tag{2} \end{equation}

So, letting $A=0$ reduces (1) and (2) to the original problems. Hope that solves the problems.

I have found, using computer, that an identity for the 9th powers is: \begin{align} 72\frac{(x^9+y^9+z^9)}{(x^3+y^3+z^3)}=&27(x^2+y^2+z^2)^2 +8(x^3+y^3+z^3)^2 \end{align}

Answer 3

Here's another way:

Let $a,b,c$ be the roots of the cubic equation $$x^3+px+q=0$$ (The $x^2$ coefficient is zero since $\sum a=0$)

Then $$\sum a^2=(\sum a)^2-2\sum ab=-2p$$ $$\sum a^3=-p\sum a-q\sum 1=-3q$$ $$\sum a^4=-p\sum a^2-q\sum a=2p^2$$ $$\sum a^5=-p\sum a^3-q\sum a^2=5pq$$ Therefore $$pq=\frac15\sum a^5=\frac13\sum a^3\cdot\frac12\sum a^2$$ Also, $$\sum a^7=-p\sum a^5-q\sum a^4=-7p^2q$$ $$\implies-p^2q=\frac17\sum a^7=(pq)(-p)=\frac15\sum a^5\cdot\frac12\sum a^2$$

Answer 4

I'll tackle the first identity. The idea is to use Newton's sums. Note that $P_2=a^2+b^2+c^2=S_1P_1-2S_2=-2S_2$. Additionally, $P_3=S_1P_2-S_2P_1+3S_3=3S_3$. Also, $$P_5=S_1P_4-S_2P_3+S_3P_2-S_4P_1+5S_5=S_3P_2-S_2P_3+5S_5$$ So we want to prove $$-S_2S_3=S_5+\frac{1}{5}(S_3P_2-S_2P_3)$$ $$-(ab+bc+ac)(abc)=\frac{1}{5}(-2abc(ab+bc+ac)-3abc(ab+bc+ac))$$ Since $S_5=0$. This is clearly true, as desired. The second can be solved similarly, but might require a bit more work.