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

Solution 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 $s_1=0$.

Solution 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}