# Chaitanya's Random Pages

## November 25, 2015

### An identity based on three numbers summing to zero

Filed under: mathematics — ckrao @ 11:06 am

Here is a nice identity which according to [1] appeared in a 1957 Chinese mathematics competition.

If $x + y + z = 0$ then

$\displaystyle \left(\frac{x^2 + y^2 + z^2}{2} \right)\left(\frac{x^5 + y^5 + z^5}{5} \right) = \left(\frac{x^7 + y^7 + z^7}{7} \right).\quad \quad (1)$

An elegant proof of this avoids any lengthy expansions. Let $x$, $y$ and $z$ be roots of the cubic polynomial

\begin{aligned} (X-x)(X-y)(X-z) &= X^3 - (x+y+z)X^2 + (xy + yz + zx)X - xyz\\ &:= X^3 + aX + b.\quad \quad (2)\end{aligned}

Then

\begin{aligned} x^2 + y^2 + z^2 &= (x+y+z)^2 - 2(xy + yz + xz)\\ &= 0 -2a\\ &= -2a\quad\quad (3)\end{aligned}

and summing the relation $X^3 = -aX - b$ for each of $X=x, X=y$ and $X=z$, gives

\begin{aligned}x^3 + y^3 + z^3 &= -a(x + y + z) - 3b\\ &= -3b.\quad\quad (4)\end{aligned}

In a similar manner, $X^4 = -aX^2 - bX$ and so

\begin{aligned} x^4 + y^4 + z^4 &= -a(x^2 + y^2 + z^2) - b(x + y + z)\\ &= -a(-2a)\\ &= 2a^2.\quad \quad (5)\end{aligned}

Next, $X^5 = -aX^3 - bX^2$ and so

\begin{aligned} x^5 + y^5 + z^5 &= -a(x^3 + y^3 + z^3) - b(x^2 + y^2 + z^2)\\ &= -a(-3b) -b(-2a)\\ &= 5ab.\quad \quad (6)\end{aligned}

Finally, $X^7 = -aX^5 - bX^4$ and so

\begin{aligned} x^7 + y^7 + z^7 &= -a(x^5 + y^5 + z^5) - b(x^4 + y^4 + z^4)\\ &= -a(5ab) -b(2a^2)\\ &= -7a^2b.\quad \quad (7)\end{aligned}

We then combine (3), (6) and (7) to obtain (1). It seems that $x^n + y^n + z^n$ for higher values of $n$ are more complicated expressions in $a$ and latex $b$, so we don’t get as pretty a relation elsewhere.

#### Reference

[1] Răzvan Gelca and Titu Andreescu, Putnam and Beyond, Springer, 2007.