Inspired by Souryajit Roy

It should be clear that $f(x^2 )$ is not a solution. For simplicity, we can check that $f(x) = x^2 - 1$ has roots $\pm 1$ , while $f(x^2) = x^4 - 1$ has roots that are $\pm 1, \pm i$ .

Many people will be tempted to select $f( \sqrt{x} )$ , since if $\alpha$ is a root of $f(x)$ , we know that $f( \alpha) = 0$ . Then $\alpha^2$ is a root of $f( \sqrt{x} )$ , since $f( \sqrt{ \alpha^2 } ) = f( \alpha ) = 0$ . However, the issue with that is that we might not necessarily end up with a polynomial. Similarly, $[ f ( \sqrt{x} ) ] ^2$ need not be a polynomial. For example, with $f(x) = x^2 + 2x + 1$ , we will get $f( \sqrt{x} ) = x + 2 \sqrt{x} + 1$ and $[f(\sqrt{x} ) ] ^2 = x^2 + 4 x \sqrt{x} + 6 x + 4 \sqrt{x} + 1$ . As such, neither of these options are valid.

The only option that remains is $f(\sqrt{x} ) f( - \sqrt{x} )$ . For completeness, we should also show that this answer satisfies the conditions of the question. It is not immediately clear that this is a polynomial (since it may suffer from issues in the previous paragraph). The easy way of seeing why this works, is that if $f(x) = \prod ( x - \alpha _ i )$ , then we are interested in finding the polynomial $g(x) = \prod ( x - \alpha_i ^2 )$ . We can obtain it as such:

$g(x) = \prod ( x - \alpha_i ^2 ) = \prod ( \sqrt{ x} - \alpha_i ) ( \sqrt{x} + \alpha_i) \\ = \prod ( \sqrt{x} - \alpha_i) \prod ( -1) ( - \sqrt{x} - \alpha_i) \\ = f( \sqrt{x} ) ( -1) ^ n f ( - \sqrt{x} ) .$

Since we do not care about the constant, we can ignore $(-1)^n$ , and conclude that $g(x) = f(\sqrt{x} ) f( - \sqrt{x} )$ is indeed the polynomial that we are looking for.

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We can easily generalize this approach, to finding polynomials where the roots have undergone a polynomial transformation (or even a rational function transformation). Give it a try: Find a polynomial whose roots are $\alpha^3 - 1$ .