Japan Mathematical Olympiad First Round

Let us try to find a polynomial $f(x)$ of degree at most $4$, which satisfies the condition of the problem with some non-zero polynomial $g(x)$. As the given equation can be rewritten in the form $f(x^3) - f(x) = (x^5 - 1)g(x)$, we see that $f(x^3) - f(x)$ must be divisible by $x^5 - 1$. Let us represent $f(x)$ as $f(x) = a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$, where $a_4, a_3, a_2, a_1, a_0$ are real numbers. Then, we get
$$f(x^3) - f(x) = a_4(x^{12} - x^4) + a_3(x^9 - x^3) + a_2(x^6 - x^2) + a_1(x^3 - x).$$
If we divide $x^{12} - x^4$, $x^9 - x^3$ and $x^6 - x^2$ by $x^5 - 1$ we get the remainders $x^2 - x^4$, $x^4 - x^3$ and $x - x^2$, respectively. Therefore, the remainder which results when we divide $f(x^3) - f(x)$ by $x^5 - 1$ is $(-a_4 + a_3)x^4 + (-a_3 + a_1)x^3 + (a_4 - a_2)x^2 + (a_2 - a_1)x$. Consequently, the fact that $f(x^3) - f(x)$ is divisible by $x^5 - 1$ is equivalent to the validity of $-a_4 + a_3 = -a_3 + a_1 = a_4 - a_2 = a_2 - a_1 = 0$. These together imply that $a_1 = a_2 = a_3 = a_4$ so that we can conclude that $f(x)$ must be of the form $f(x) = ax^4 + ax^3 + ax^2 + ax + b$, with some pair of real numbers $a, b$. When this is the case, we can determine from $f(x^3) - f(x) = (x^5 - 1)g(x)$ that $g(x) = ax^7 + ax^4 + ax^2 + ax$. From the fact that neither $f(x)$ nor $g(x)$ are identically zero, we get that the constant $a$ must be non-zero and $b$ can be arbitrary. Hence we see that $(f(x), g(x)) = (ax^4 + ax^3 + ax^2 + ax + b, ax^7 + ax^4 + ax^2 + ax)$ gives a pair $(f(x), g(x))$, satisfying the given functional equation. Since any polynomial of degree $4$ or less can be represented in the form $a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$ for some choice of real numbers $a_4, a_3, a_2, a_1, a_0$, the argument above shows that the polynomial $f(x)$ desired must be of the form $f(x) = ax^4 + ax^3 + ax^2 + ax + b$ for some choice of non-zero real number $a$ and a real number $b$.

Alternate Solution:
We can use the following theorem from the complex function theory (de Moivre's Theorem):
For an arbitrary angle $\theta$ and a positive integer $n$
$$(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta$$
holds. Here $i$ represents the imaginary unit $\sqrt{-1}$.
Now, if we let $\theta = \frac{2\pi}{5}$ and set $\omega = \cos\theta + i\sin\theta$, then by de Moivre's Theorem, we get $\omega^5 = 1$. For any positive integer $k$, we also have $(\omega^k)^5 = (\omega^5)^k = 1$ so that each of the five numbers $1, \omega, \omega^2, \omega^3, \omega^4$ gives a complex-valued root to the equation $x^5 - 1 = 0$. From the fact that $\omega^k = \cos k\theta + i \sin k\theta$, it follows that these five numbers are all distinct, and therefore by the factorization theorem, we obtain that
$$x^5 - 1 = (x - 1)(x - \omega)(x - \omega^2)(x - \omega^3)(x - \omega^4).$$
Dividing both sides of the equation above by $x - 1$, we get
$$x^4 + x^3 + x^2 + x + 1 = (x - \omega)(x - \omega^2)(x - \omega^3)(x - \omega^4).$$
If we substitute $x = \omega, \omega^2, \omega^3, \omega^4$ into the identity $f(x^3) - f(x) = (x^5 - 1)g(x)$, we obtain, since $x^5 - 1 = 0$,
$$f(\omega^3) = f(\omega), \quad f(\omega^6) = f(\omega^2), \quad f(\omega^9) = f(\omega^3), \quad f(\omega^{12}) = f(\omega^4),$$
respectively. Since $\omega^6 = \omega$, $\omega^9 = \omega^4$, $\omega^{12} = \omega^2$, we see that the identities $f(\omega) = f(\omega^2) = f(\omega^3) = f(\omega^4)$ hold. Let $b$ denote the common value of these quantities, then since $f(\omega) - b = f(\omega^2) - b = f(\omega^3) - b = f(\omega^4) - b = 0$, we can conclude from the factorization theorem that $f(x) - b$ is divisible by
$$(x - \omega)(x - \omega^2)(x - \omega^3)(x - \omega^4) = x^4 + x^3 + x^2 + x + 1.$$
Therefore, there exists a polynomial $h(x)$ such that $f(x) = h(x)(x^4 + x^3 + x^2 + x + 1) + b$.
If $h(x) = 0$, we get $f(x^3) - f(x) = 0$, which forces $g(x) = 0$, contradicting the assumption. Therefore, we must have $h(x) \neq 0$, and this shows that the degree of $f(x)$ must be at least $4$. When $h(x)$ is a non-zero constant function, then the corresponding $f(x)$ has the minimal degree of $4$, and if we write $h(x) = a$ for some non-zero number $a$, we get
$$(f(x), g(x)) = (a(x^4 + x^3 + x^2 + x) + b, a(x^7 + x^4 + x^2 + x))$$
as before.