MOPSS, 22nd November 2025

Note that \[ (1 + a^2)(1 + b^2) \geq (1 + ab)^2 \] holds for any real numbers \(a,b\) by the Cauchy–Schwarz inequality. This implies that

\(\begin{align*} & \frac{1 + x_1^2}{1 + x_1 x_2} + \frac{1 + x_2^2}{1 + x_2 x_3} + \cdots + \frac{1 + x_n^2}{1 + x_n x_1} \\ & \geq \sqrt{\frac{1 + x_1^2}{1 + x_2^2}} + \sqrt{\frac{1 + x_2^2}{1 + x_3^2}} + \cdots + \sqrt{\frac{1 + x_n^2}{1 + x_1^2}} \\ & \geq n , \end{align*} \\)

where the last inequality follows from the AM–GM inequality. This completes the proof.

Without loss of generality, let us consider the case that the equations \[ (x - a)(x - b) = x - c, (x - b)(x - c) = x - a \] do not any real root. It follows that their discriminants are negative, that is, \[ (a + b + 1)^2 - 4 (ab + c), (b + c + 1)^2 - 4(bc + a) \] are negative. Note that

\(\begin{align*} (a + b + 1)^2 - 4(ab + c) & = (a - b)^2 + 1 + 2(a + b - 2c)\\ & = (b - a + 1)^2 + 4(a - c),\\ (b + c + 1)^2 - 4(bc + a) & = (b - c)^2 + 1 + 2(b + c - 2a)\\ & = (b - c + 1)^2 + 4(c - a). \end{align*} \\)

Since \(a\) and \(c\) are distinct real numbers, one of \(a - c\) and \(c - a\) is positive. Thus, the discriminant of one of the equations \[ (x - a)(x - b) = x - c, (x - b)(x - c) = x - a \] is non-negative, which is a contradiction. Therefore, at least two of the three equations have real solutions.

Note that the sum of the discriminants of the quadratic polynomials \[ x^2 + (a - b)x + (b - c), x^2 + (b - c)x + (c - a), x^2 + (c - a)x + (a - b) \] are equal to

\(\begin{align*} & (a - b)^2 - 4(b - c) + (b - c)^2 - 4(c - a) + (c - a)^2 - 4(a - b) \\ & = (a - b)^2 + (b - c)^2 + (c - a)^2 , \end{align*} \\)

which is non-negative, and hence, at least one of the discriminants is non-negative. This implies that at least one of the three equations has a real root.

Let \(p\) be a prime number, and let \(n\) be the smallest positive integer, strictly greater than 1, for which \(n^6 - 1\) is divisible by \(p\). Let us consider the case that \(p\) divides \(n^2 - n + 1\). Note that \[ n^2 - n + 1 = (n - 1)^2 + (n - 1) + 1 \] holds. Then \(p\) divides \((n - 1)^2 + (n - 1) + 1\), implying that \(p\) divides \((n - 1)^6 - 1\). Using the minimality of \(n\), it follows that \(n - 1\) is equal to \(1\). This implies that \(p\) divides \(2^2 - 2 + 1 = 3\). This shows that \(p\) is equal to \(3\). Using \(p = 3\) and \(n = 2\), it follows that \((n + 2)^6 - 1\) is divisible by \(p\). Now, let us consider the case that \(p\) does not divide \(n^2 - n + 1\). If \(p = 2\), then \(p\) divides \((n + 2)^6 - 1\). Henceforth, let us assume that \(p\) is an odd prime. Since \(p\) divides \((p - 1)^6 - 1\), using the minimality of \(n\) and that \(p - 1 \geq 2\), it follows that \(n\) is less than or equal to \(p - 1\). If \(n\) is less than \(p - 1\), then using the factorization \[ n^6 - 1 = (n - 1)(n + 1)(n^2 - n + 1)(n^2 + n + 1), \] it follows that \(p\) divides \(n^2 + n + 1\), and as consequence, \(p\) divides \((n + 1)^2 - (n + 1) + 1\), which is a divisor of \((n + 1)^6 - 1\). Finally, if \(n\) is equal to \(p - 1\), then \(p\) divides \((n + 2)^6 - 1\). This completes the proof.