# Given $x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,$ can we conclude that $x^n+y^n+z^n\in\mathbb{Q}$ for all $n\in\mathbb{N}$. Is there an explicit form?

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 $$x^n+y^n+z^n$$.

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

$$T^3-T^2-\frac{1}{2}T-\frac{1}{6}$$

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