Yesterday, I wrote a post on the Mason-Stothers theorem and presented an elementary proof of the theorem given by Noah Snyder. As mentioned in that post, I will present now a problem proposed by Magkos Athanasios (Kozani, Greece) that can be solved almost “effortlessly” using the aforesaid theorem.

Problem: Let $f$ and $g$ be polynomials with complex coefficients and let $a \ne 0$ be a complex number. Prove that if $(f(x))^3 = (g(x))^2 + a$

for all $x \in \mathbb{C}$, then the polynomials $f$ and $g$ are constants.

(Magkos Athanasios)

Solution: First, note that if $f$ is a constant, then this forces $g$ to be a constant, and vice-versa. Now, suppose $f$ and $g$ are not constants. We show that this leads to a contradiction.

Observe that if $f$ and $g$ have a common root, say, $\alpha$, then we have $(f(\alpha))^3 = (g(\alpha))^2 + a$, which implies $0 = 0 + a$, which implies $a = 0$, a contradiction. Therefore, we conclude $f, g$ and $a$ are relatively prime polynomials, and hence, $f^3, g^2$ and $a$ are also relatively prime. Now, let $\deg (f) = n$ and $\deg (g) = m$. Then, from the given equation, we conclude $n = 2k$ and $m = 3k$ for some $k \in \mathbb{N}$.

So, $\max \{\deg(f^3), \deg(-g^2), \deg(a)\} = \max \{6k, 6k, 0\} = 6k$.

Also, $N_0 (f^3 (-g^2) a) - 1$ $= N_0 (fg) - 1 \le \deg (f) + \deg(g) - 1 = 2k + 3k - 1 = 5k - 1$.

Now, applying the Mason-Stothers theorem, we get $6k \le 5k - 1$, which implies $k \le -1$, a contradiction! And, we are done.