MOPSS, 8th November 2025

On the contrary, let us assume that none of the numbers on the circle is divisible by \(3\). Then, each number on the circle is congruent to either \(1\) or \(2\) modulo \(3\). The given conditions imply that any two neighbouring numbers on the circle are not congruent modulo \(3\). Since there are \(99\) numbers on the circle and \(99\) is odd, it is impossible to have such a configuration.

Denote the common difference of the arithmetic progression by \(d\). Since \(p, q\) are distinct primes, it follows that \(p\) does not divide \(p^{23} - q^{23}\). Using that \(d\) divides \(p^{23} - q^{23}\), we conclude that \(p\) does not divide \(d\). Note that \(d\) also divides \(p^{24} - p^{23}\), which implies that \(d\) divides \(p - 1\). It follows that \(d\) divides \(p^{23} - p\). Since \(d\) is negative, we conclude that \(p\) occurs in the arithmetic progression. By a similar argument, it follows that \(q\) also occurs in the arithmetic progression.

Note that if \(x, y, z\) are rational numbers satisfying the given equation, then \(dx, dy, dz\) also satisfy the equation for any positive integer \(d\). Hence, it suffices to prove that there are no integer solutions to the given equation other than the trivial solution \(x = y = z = 0\). We claim that if \((a, b, c)\) are integers satisfying \[ x^3 + 3y^3 + 9z^3 = 9xyz, \] then \(3\) divides \(a\), and \((b, c, a/3)\) also satisfies the above equation. Indeed, if we assume that if \((a, b, c)\) are integers satisfying \[ x^3 + 3y^3 + 9z^3 = 9xyz, \] then note that \(3\) divides \(a^3\). Since \(3\) is a prime, it follows that \(3\) divides \(a\). Using \[ a^3 + 3b^3 + 9c^3 = 9abc, \] we obtain \[ b^3 + 3c^3 + 9 \left( \frac{a}{3} \right)^3 = 9 bc \left( \frac{a}{3} \right) . \] This completes the proof of the claim.

Let \((x, y, z)\) be a non-trivial integer solution to the given equation with \(\vert x \vert + \vert y \vert + \vert z \vert\) minimum. Note that \(x\) is nonzero, otherwise, \(y, z\) satisfy \(y^3 + 3 z^3 = 0\), which is impossible since \(y, z\) are integers. By the above claim, it follows that \(y, z, x/3\) is also a solution to the given equation. Using that \(x\) is nonzero, we obtain \[ \vert y \vert + \vert z \vert + \left \vert \frac{x}{3} \right \vert < \vert x \vert + \vert y \vert + \vert z \vert , \] which contradicts the minimality of \(\vert x \vert + \vert y \vert + \vert z \vert\). This shows that there are no non-trivial integer solutions to the given equation.

Note that it suffices to consider the case when \(r\) is the power of a prime, say \(r = p^a\) for some prime \(p\), and some positive integer \(a\). Indeed, if \(x\) is a prime power dividing \(r\), then for each positive integer \(k\), we have \[ d(k s) \geq d(k r) \geq d(k x) , \] and if we can show that every prime power dividing \(r\) also divides \(s\), then it follows that \(r\) divides \(s\). Observe that \(s > 1\) since \(d(s) \geq d(p^a) \geq 2\). Write \[ s = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k} , \] where \(k\) is a positive integer, \(p_1, p_2, \dots, p_k\) are distinct primes, and \(\alpha_1, \alpha_2, \dots, \alpha_k\) are positive integers. Let us first show that \(p = p_i\) for some \(i \in {1, 2, \dots, k}\). Suppose, for the sake of contradiction, that \(p \neq p_i\) for all \(i \in {1, 2, \dots, k}\). Then, taking \(k = p^u\) with \[ u = a(\alpha_1 + 1)(\alpha_2 + 1) \dots (\alpha_k + 1) - a , \] we have

\(\begin{align*} d(ks) & = (u + 1) (\alpha_1 + 1)(\alpha_2 + 1) \dots (\alpha_k + 1) ,\\ d(kr) & = d(p^{u + a})\\ & = u + a + 1 . \end{align*} \\)

Since \(u + a + 1\) is coprime to \((\alpha_1 + 1)(\alpha_2 + 1) \dots (\alpha_k + 1)\), it follows that \(u + a + 1\) divides \(u + 1\). This is a contradiction since \(a \geq 1\). Thus, \(p = p_i\) for some \(i \in {1, 2, \dots, k}\). Renaming indices if necessary, we may assume that \(i = 1\). Taking \(k = p^{u}\) with \[ u = a(\alpha_1 + 1)(\alpha_2 + 1) \dots (\alpha_k + 1) - a , \] we have

\(\begin{align*} d(ks) & = (u + \alpha_1 + 1) (\alpha_2 + 1) \dots (\alpha_k + 1) ,\\ d(kr) & = d(p^{u + a})\\ & = u + a + 1 . \end{align*} \\)

Since \(u + a + 1\) is coprime to \((\alpha_2 + 1) \dots (\alpha_k + 1)\), it follows that \(u + a + 1\) divides \(u + \alpha_1 + 1\). This shows that \(a \leq \alpha_1\).

Note that for any \(x \in \mathbb{R}\), we have \[ f(x + 1)^2 \leq f(x) , \] which shows that \(f(x) \geq 0\). To show that \(f(x) \leq 1\) for any \(x \in \mathbb{R}\), let us establish the following claim.

For any \(x, y \in \mathbb{R}\) with \(x > y\), and for any positive integer \(n\), the inequality \[ f(x)^{2^n} \leq f(y) \] holds.

Proof of the claim. Note that the inequality holds for \(n = 1\) by hypothesis. Let us assume that \(n\) is a positive integer such that the inequality holds for any \(x, y \in \mathbb{R}\) with \(x > y\). Then, for any \(x, y \in \mathbb{R}\) with \(x > y\), we have \[ x > \frac{x + y}{2} > y , \] which implies that \[ f(x)^{2^n} \leq f\left(\frac{x + y}{2}\right) \quad \text{and} \quad f\left(\frac{x + y}{2}\right)^{2} \leq f(y) . \] Combining these two inequalities and using that \(f(x)\) is nonnegative, we obtain \[ f(x)^{2^{n + 1}} \leq f(y) . \] By the principle of mathematical induction, the claim follows. Let \(x\) be an arbitrary real number. Using the above claim, we obtain \[ f(x)^{2^n} \leq f(x - 1) \] for any positive integer \(n\). If \(f(x) \leq 1\), then we are done. It remains to consider the case that \(f(x) > 1\), which we assume from now on. It follows that \[ f(x - 1) \geq 1 + 2^n \left(f(x) - 1\right) \] for any positive integer \(n\). This is impossible since \(f(x) - 1\) is positive. Combining the above, we conclude that \[ 0 \leq f(x) \leq 1 \] holds for any \(x \in \mathbb{R}\).