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}.


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


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.