I had seen a fun problem that is exactly the problem in the question, except it was a specific case of this. Turns out, if $x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,$ then $x^5+y^5+z^5 = 6$. I put this equation into Wolfram Alpha for different values of $n$. For instance, $x^8+y^8+z^8=\frac{51}{72}$. $x^{11}+y^{11}+z^{11}=\frac{11117}{216}.$

Firstly, is there an explicit expression to evaluate $x^n+y^n+z^n$? And, if not, are we able to prove that this is always a rational number, at least? WolframAlpha is unable to calculate an explicit form for $x^n+y^n+z^n$.


2 Answers 2


None of the calculations we need are mysterious, but one might need some experience dealing with symmetric functions and their relations to know what to look for.

We denote $x^n+y^n+z^n$ by $p_n$. You can verify the following inductive formula:

$p_n = (x+y+z)p_{n-1}-(xy+yz+zx)p_{n-2}+(xyz)p_{n-3}$

We can calculate the coefficients of the $p_i$ above as follows:

$x+y+z = 1$

$xy + yz + zx = \frac{(x+y+z)^2 -(x^2+y^2+z^2)}{2} = \frac{1^2-2}{2} = -\frac{1}{2}$

$xyz = \frac{(x^3+y^3+z^3)-(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{3} = \frac{3-1 \cdot (2-(-\frac{1}{2}))}{3} = \frac{1}{6}$

Thus our inductive formula becomes

$p_n = p_{n-1}+\frac{1}{2}p_{n-2}+\frac{1}{6}p_{n-3}$

It is clear from this recursive equation that all of the $p_n$ will be rational numbers, since $p_1, p_2$ and $p_3$ are rational numbers. In fact, it even follows that the denominator of $p_n$ will be of the form $2^a3^b$ for some non-negative integers $a$ and $b$. For an explicit form we could look at roots of the characteristic/auxiliary polynomial of the corresponding recursive equation. In this case we get


whose roots are not that nice. However, Wolfram Alpha gives an explicit formula. Then again, one could perhaps just solve the original system directly to find $x$, $y$ and $z$.


Yes. In https://en.m.wikipedia.org/wiki/Symmetric_polynomial, look at Power-sum Symmetric Polynomials in the section Special kinds of symmetric polynomials


This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .