<s>[INST] 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
 [/INST] $2(xy+xz+yz)=(x+y+z)^2-(x^2+y^2+z^2)\in\mathbb Z$ 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!
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