India_2017

We can easily extend the definition of a square polynomial to polynomials with complex coefficients. In all the arguments below we consider polynomials with complex coefficients.

Lemma: If $p(x)$ is a square and $\alpha$ is a non-zero complex number then $p(x)-\alpha$ is not a square.

Proof of Lemma: Suppose $p(x)=q(x{)}^2$ and $p(x)-\alpha =r(x{)}^2$, with both $q(x)$ and $r(x)$ being non-constant polynomials. Then $\alpha =(q(x)-r(x))(q(x)+r(x))$. Clearly, either $q(x)-r(x)$ or $q(x)+r(x)$ is not a constant polynomial, and hence a contradiction. This proves the lemma.

Continuation of the solution: We can write $f(x)$ as $f_1(x{)}^2(x-{\alpha }_1)(x-{\alpha }_2)\cdots (x-{\alpha }_k)$, where $f_1(x)$ is a polynomial and ${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_k$ are distinct complex numbers. Then $f(g(\tau ))=f_1(g(\tau ){)}^2(g(\tau )-{\alpha }_1)\cdots (g(\tau )-{\alpha }_k)$ is a square. It follows that $(g(x)-{\alpha }_1)(g(x)-{\alpha }_2)\cdots (g(x)-{\alpha }_k)=h(x{)}^2$ for some polynomial $h(x)$. Let $\beta$ be such that $g(\beta )={\alpha }_1$. Then $h(\beta )=0$ and hence $h(x)=(x-\beta )h_1(x)$. Note that $g(\beta )-{\alpha }_i=0$ for any $i=1$. Therefore it follows that $(x-\beta {)}^2$ divides $g(x)-{\alpha }_1$. We can then conclude that $g(x)-{\alpha }_1$ is a square. Similarly, $g(x)-{\alpha }_i$ is a square for $i=1,2,\cdots ,k$. By the above lemma, it follows that $k=1$, so $f(x)=f_1(x{)}^2(x-\alpha )$ and $g(x)=g_1(x{)}^2+\alpha$ for some non-zero complex number $\alpha$. Therefore $g(f(x))=g_1(f(x){)}^2+\alpha$ and hence by the above lemma it follows that $g(f(x))$ is not a square. This completes the proof.