Let $P(x)$ be a polynomial with integer coefficients and $deg(P(x))>0$. Is it true that if $P(x)$ is a square number for every integers, then $P(X)=(Q(x))^2$, with $Q(x)$ being a polynomial? Is the statement correct if $P(x)$ is a square number for infinite integers, not every integers ?
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$P(x) = x$ is a perfect square for infnitely many integer inputs, but it is not the square of any other polynomial. So the answer to your second question is no. – Arthur Oct 29 '18 at 13:25
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You don't need the condition $\deg(P(x))>0$. If $P(x)=a^2$ is constant, then you can take $Q(x)=a$. – TonyK Oct 29 '18 at 13:27
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1An even stronger statement. – metamorphy Oct 29 '18 at 13:31
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Special case of a well-known result of Davenport, Lewis and Schinzel cited in this answer. – Bill Dubuque Oct 29 '18 at 14:59