show that $xyz\in N$ if $x^n+y^n+z^n\in Z$ Let $x,y,z\in R$ and such for any postive integers $n$ have
$$a_{n}=x^n+y^n+z^n\in Z$$
show that $xyz\in Z$
I have use
$$a_{n+3}=(x+y+z)a_{n+2}-(xy+yz+xz)a_{n+1}+xyza_{n}$$
since $2(xy+yz+xz)=(x+y+z)^2-(x^2+y^2+z^2)\in Z$
but $xy+yz+xz$ can't integers,because $2(xy+yz+xz)$ is integers
 A: $2(xy+xz+yz)=(x+y+z)^2-(x^2+y^2+z^2)\in\mathbb Z$ and since
$$(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(xy+xz+yz)-3xyz,$$ we see that $6xyz\in\mathbb Z$.
Now, $$x^4+y^4+z^4=$$
$$=(x+y+z)^4-4(x+y+z)^2(xy+xz+yz)+2(xy+xz+yz)^2+4(x+y+z)xyz$$ or
$$3(x^4+y^4+z^4)=$$
$$=3(x+y+z)^4-12(x+y+z)^2(xy+xz+yz)+6(xy+xz+yz)^2+12(x+y+z)xyz,$$
which says $6(xy+xz+yz)^2\in\mathbb Z$ and from here $xy+xz+yz\in\mathbb Z$.
Thus, $3xyz\in\mathbb Z$ and $4(x+y+z)xyz\in\mathbb Z$.
Let $xyz\in\mathbb Z$ is wrong. 
Then $x+y+z$ divided by $3$.
But
$$x^5+y^5+z^5=$$
$$=(x+y+z)^5-5(x+y+z)^3(xy+xz+yz)+5(x+y+z)(xy+xz+yz)^2+$$
$$+5((x+y+z)^2-(xy+xz+yz))xyz,$$
which says that $5((x+y+z)^2-(xy+xz+yz))xyz\in\mathbb Z$, 
which gives $xy+xz+yz$ divided by $3$.
Now, from $$x^6+y^6+z^6=$$
$$=(x+y+z)^6-6(x+y+z)^4(xy+xz+yz)+9(x+y+z)^2(xy+xz+yz)^2-$$
$$-2(xy+xz+yz)^3+(6(x+y+z)^3-4(x+y+z)(xy+xz+yz))xyz+3x^2y^2z^2$$ we obtain $3x^2y^2z^2\in\mathbb Z$, which is contradiction.
Thus, $xyz\in\mathbb Z$ and we are done!
A: Here is an approach using algebraic number theory:
Lemma. $\{x_i\}_{i=1}^{m}$ are complex numbers such that $\sum_i x_i^k$ are  integers for every $k \in \Bbb N^{*}$, then $x_i$ are all algebraic integers.
Proof: 
Firstly, by Newton identity we know $x_i$ are all algebraic numbers. ($\sum_i x_i^k$  $\in \Bbb Z ,\forall k \Rightarrow \sigma_k(x_i) \in \Bbb Q $ ,$\forall k$, where $\sigma_k$ is the k-th elementary symmetric polynomial).
Pick a number field $K$ contains all $x_i$, for every prime valuation v over K, we have $|\sum_i x_i^k|_v \leq 1, \forall k$ as $\sum_i x_i^k$ are algebraic integers.
Consider $ f(t)=\sum_{i=1}^m\frac{1}{1-t{x_i}}$ over $\Bbb C_p$, where $p$ is the restriction of $v$ to $\Bbb Q$. It converges for $|t|_v<1$ by above esitimates 
$|\sum_i x_i^k|_v \leq 1, \forall k$, hence has no poles in the disc $\{t \in \Bbb C_p ||t|_v<1\}$, therefore $|x_i|_v \leq 1, \forall i=1,\dots,m$
So $x_i \in \bigcap_{v} O_{K_v}$ , hence must lies in $O_K$.
In sum, we show $x_i$ are all algebraic integers.
Come back to your question: 
By Newton identity we know that $\prod_{i}x_i \in \mathbb Q$, but $x_i \in O_K$, so $\prod_{i}x_i \in \mathbb Q \cap O_K= \mathbb Z$. The same things hold for other elementary symmetric polynomials.
