I have following problem, that I know that is true, but I don't know how to prove.
For all polynomial with integer coefficient, if the polynomial gives perfect square value for all positive integer, then polynomial itself is square, that is, $P(x)=R(x)^2 $ for some integer coefficient polynomial $R(x)$
Well, it is natural to think of the following question.
Q. For all positive integers, $P(x)$ gives perfect $k$-power. is $P(x)=R(x)^k $ for some integer coefficent polynomial $R(x)$?
I assume that it's true. Even though I don't know how to prove it. It's not what I want to talk about now.
Q. For infinitely many positive integers, $P(x)$ gives square value. is $P(x)=R(x)^2 $ for some integer coefficent polynomial $R(x)$?
The answer is false, hence one can think of $y^2=2x^2+1$, the equation has infinitely many integer solutions, thus being a counterexample.
However, once one notice that density of the solution is $O(\log x)$, (for which, above case, it is.) One also get the question of having dense enough solution, polynomial must be perfect power itself.
so here is my third question.
Q. Denote the number of solution to the following equation as $C(x)$. if $C(x)>>x,$ then $P(x)=R(x)^2 $for some integer coefficient polynomial $R(x)$.
Does this kindof theorem, topic, has a name? Is this already well-known? Thank you.
if $P$ is irreducible :
step 1:$(P(x),P'(x))=1$ so there are polynomial $Q,R\in \mathbb{Q}$ such that $QP+RP'=1$ if n was the g.c.d of denominators of the coefficients of $Q,R$ then $(nQ(m))P(m)+(nR(m))P'(m)=n$ so for each $m\in N$ , $(P(m),P'(m))|n$
step 2:$P(x+kq)=P(x)+kqP'(x) \hspace{0.25 cm} mod q^2$ if $q|P(x)$ and P(x) was a complete power relatively prime to $n$ then $q|P(x+kq)$ but $q^2\not |P(x+kq) $
so you can find some bounds on $C_P(x)$ in your problem.
general case:
hint:if $P(x)=Q(x)S(x)$ and $(Q,S)=1$ use the same arguments as the step one to relate the $C_P(x)$ to $C_Q(x),C_S(x)$