Japan Mathematical Olympiad

The assertion of the problem is obvious if $P(x)$ is a constant. Therefore, we assume in the sequel that $P(x)$ is a polynomial of degree $m$, where $m$ is a positive integer. As $P(x)$ is a polynomial with integer coefficients and satisfies $P(n^2)=0$, there exists, by the factorization theorem, a polynomial $Q(x)$ with integer coefficients such that $$P(x)=Q(x)(x-n^2)$$ holds. Suppose that $P(a^2)=1$ holds for some nonzero rational number $a$. We can represent $a$ as $a=\frac{q}{p}$ with $p$ being a positive integer, $q$ a nonzero integer and $p,q$ relatively prime. Therefore, if we substitute $x=a^2$ in the equation above, we obtain $$P(a^2)=Q(a^2)(a^2-n^2)=\frac{1}{p^{2m}}\left(p^{2(m-1)}Q\left({\left(\frac{q}{p}\right)}^2\right)\right)(q^2-(pn{)}^2),$$ from which we conclude as $P(a^2)=1$ that $$\left(p^{2(m-1)}Q\left({\left(\frac{q}{p}\right)}^2\right)\right)(q^2-(pn{)}^2)=p^{2m}.$$ We note that both $p^{2(m-1)}Q\left({\left(\frac{q}{p}\right)}^2\right)$ and $q^2-(pn{)}^2$ are integers, and therefore, the validity of the identity above implies that $q^2-(pn{)}^2$ must be a factor of $p^{2m}$. However, since $p,q$ are relatively prime, neither $q-pn$ nor $q+pn$ has a common prime factor with $p$. Furthermore, since none of $n,p,q$ is zero, it is easy to check that it is impossible for $q-pn$ and $q+pn$ to have absolute value 1 simultaneously. Putting these facts together we conclude that neither $q-pn$ and $q+pn$ can be a factor of $p^{2m}$, and hence $q^2-(pn{)}^2=(q-pn)(q+pn)$ cannot be a factor of $p^{2m}$, contradicting the conclusion obtained above from the assumption $P(a^2)=1$, and this contradiction proves the assertion of the problem.