Let $P$ and $Q$ be polynomials with complex coefficients such that $P(P(x)) = Q(Q(x))$. Prove that $P = Q$.
It is obvious that degree of both will be equal. But I don't have any idea how to solve this question.
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1Perhaps you gave forgotten an assumption on $p$ and $q$ - like that they are monic polynomials ? – Dietrich Burde Dec 27 '15 at 09:26
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This is clearly false. For instance, $P(x) = -x$ and $Q(x) = x$. We then have $$P(P(x)) = x = Q(Q(x))$$
There are many other counter examples as well. For instance, even in the case of linear functions, there are infinite counter example. For instance, if $P(x) = ax+b$ and $Q(x) = cx+d$, we then have $$P(P(x)) = P(ax+b) = a(ax+b)+b = Q(cx+d) = c(cx+d) + d$$ This means we need $$a^2x + b(a+1) = c^2x+d(c+1)$$ This means $a^2=c^2$ and $b(a+1) = d(c+1)$. If $a=c$ and $a,c\ne -1$, then $b=d$. If $a=-c$, then $d=b \cdot \dfrac{1+a}{1-a}$.
Hence, for instance, $c=-2$, $a=2$, $d=-3b$. Hence, this gives $P(x) = 2x+b$ and $Q(x) = -2x-3b$.
This gives us $$P(P(x)) = P(2x+b) = 2(2x+b) + b = 4x+3b$$ and $$Q(Q(x)) = Q(-2x-3b) = -2(-2x-3b)-3b = 4x+3b$$
