<s>[INST] If $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$ 
$P,Q,R,S$ are polynomials such that: $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$

I thought a lot on this but no result!!
By the way,one idea is to insert some values for $x$ and try to produce a system of equations for the given polynomials,but I'm not sure it works.
 [/INST] Firstly it is easy to see that the roots of $x^4+x^3+x^2+x+1 =0$ are all different. Also if $a$ is a root for $ x^4+x^3+x^2+x+1=0$ then $a^5=1$. The same holds for the rest of three roots: $b,c,d$.
Now if we plug in each root in
$P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ 
we get:
$$P(1)+aQ(1)+a^2R(1)= 0$$
$$P(1)+bQ(1)+b^2R(1)= 0$$
$$P(1)+cQ(1)+c^2R(1)= 0$$
$$P(1)+dQ(1)+d^2R(1)= 0$$
So quadratic function $f(x) = P(1)+xQ(1)+x^2R(1)$ has 4 different roots and so $f(x)=0$ for all $x$, so $P(1)=Q(1)=R(1)=0$ and thus the conclusion.
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