MOPSS, 11th October 2025

Note that \[ \frac{1}{2n} = \left( \frac{1}{n} - \frac{1}{n + 1}\right) + \left( \frac{1}{n + 1} - \frac{1}{n + 2} \right) + \dots + \left( \frac{1}{2n + 1} - \frac{1}{2n}\right) \] holds. This yields that \(\frac{1}{n}\) is the sum of the reciprocals of \[ \frac{n(n + 1)}{2}, \frac{(n + 1)(n + 2)}{2}, \dots, \frac{(2n + 1)2n}{2} , \] which are distinct triangular numbers since the map \(x \mapsto \frac{x(x + 1)}{2}\) from \(\mathbb{R} \to \mathbb{R}\) is injective.

Let \(p\) be a prime number such that \(p\), \(p + 2\), \(p + 6\), \(p + 8\) and \(p + 12\) are all prime. Note that \(p\) is odd. Since \(p + 6, p + 8\) are primes greater than \(3\), it follows that \[ p \not\equiv 0 \pmod{3}, \quad p \not\equiv 1 \pmod{3}, \] which implies that \(p \equiv 2 \pmod{3}\). Since \(p + 6, p + 8, p + 12\) are primes greater than \(5\), it follows that \[ p \not \equiv 4 \pmod{5}, \quad p \not \equiv 2 \pmod{5}, \quad p \not \equiv 3 \pmod{5} , \] which implies that \(p = 5\) or \(p \equiv 1 \pmod{5}\). Note that if \(p = 5\), then \(p + 2, p + 6, p + 8, p + 12\) are all prime. Henceforth, let us assume that \(p \equiv 1 \pmod{5}\). Note that \(p + 2, p + 6, p + 8, p + 12\) are all primes greater than \(7\). It follows that \(p\) not congruent to any of \(5, 1, 6, 2\) modulo \(7\). This shows that \(p = 7\) or \(p\) is congruent to either \(3\) or \(4\) modulo \(7\). Note that if \(p = 7\), then \(p + 2\) is not a prime. This implies that \(p\) is congruent to either \(3\) or \(4\) modulo \(7\). Since \(p \equiv 2 \pmod{3}\) and \(p \equiv 1 \pmod{5}\), it follows that \(p\) is congruent to \(11\) modulo \(15\). If \(p \leq 200\), then \(p\) is equal to one of the integers \[ 11, 41, 71, 101, 131, 161, 191. \] Among the above integers, only \(11, 101\) satisfy the required congruence condition modulo \(7\). Note that if \(p = 11\), then \(p + 2, p + 6, p + 8, p + 12\) are all prime. Moreover, if \(p = 101\), then \(p + 2, p + 6, p + 8, p + 12\) are all prime. This shows that there are precisely three prime numbers \(p\) less than \(200\) for which \(p + 2, p + 6, p + 8, p + 12\) are all prime, and these are \(5, 11, 101\).

Let \(q\) be a prime number such that \(q, q + 2, q + 6, q + 8, q + 12, q + 14\) are all prime. Since \(q + 6, q + 8\) are primes greater than \(3\), it follows that \(q \equiv 2 \pmod{3}\). Note that \(q\) is odd, and hence, \(q\) is at least \(5\). Since \(q + 6, q + 8, q + 12, q + 14\) are primes greater than \(5\), it follows that \(q\) is equal to \(5\). Note that if \(q = 5\), then \(q + 2, q + 6, q + 8, q + 12, q + 14\) are all prime. This shows that there is only one prime number \(q\) for which \(q + 2, q + 6, q + 8, q + 12, q + 14\) are all prime, and this is \(5\).

Let \(m, n\) be positive integers satisfying \[ n^2 - 6n = m^2 + m - 10. \] This yields \[ (n - 3)^2 = m^2 + m - 1, \] which implies \[ (n - 3)^2 = \left( m + \frac{1}{2} \right)^2 - \frac{5}{4}. \] Rearranging gives \[ (2m + 1)^2 - (2(n - 3))^2 = 5 . \] Factorizing yields \[ (2m + 1 - 2(n - 3)) (2m + 1 + 2(n - 3)) = 5 . \] Using \(n^2 - 6n = m^2 + m - 10\), it follows that \(n\) is not an odd integer. Note that if \(n = 2\), then the given equation holds yields \(m = 1\). It remains to consider the case that \(n \geq 4\), which we assume from now on. Note that \[ 2m + 1 - 2 (n - 3) \leq 2m + 1 + 2 (n - 3) \] holds, and hence we obtain \[ 2m + 1 - 2(n - 3) = 1, \quad 2m + 1 + 2(n - 3) = 5. \] This gives \(m = 1, n = 4\).

Moreover, if \(m = 1, n = 4\), then \(n^2 - 6n = m^2 + m - 10\) holds. Further, if \(m = 1, n = 2\), then also the given equation holds.

Therefore, the pairs \((m, n)\) of positive integers which satisfy the equation \(n^2 - 6n = m^2 + m - 10\) are precisely \((1, 2)\) and \((1, 4)\).

Let \(n\) be a positive integer such that \(n \times 2^n + 1 = k^2\) for some positive integer \(k\). Rearranging gives \(n \times 2^n = k^2 - 1 = (k - 1)(k + 1)\). Since the product \((k - 1)(k + 1)\) of the consecutive integers \(k - 1\) and \(k + 1\) is even, it follows that \(k - 1, k + 1\) are both even. Thus, we can write \(k - 1 = 2a\) and \(k + 1 = 2b\) for some positive integers \(a\) and \(b\) with \(b = a + 1\). Note that the greatest common divisor of the integers \(k - 1\) and \(k + 1\) is \(2\). Therefore, the greatest common divisor of the integers \(a\) and \(b\) is \(1\). This shows that at least one of the integers \(a\) and \(b\) is odd. Consequently, one of the integers \(k - 1\) and \(k + 1\) is divisible by \(2^{n - 1}\). This shows that \[ n \times 2^n \geq 2^{n - 1} (2^{n - 1} - 2) \] holds, which simplifies to \(n \geq 2^{n - 2} - 1\). Using induction, it follows that \(2^{m - 2} > m + 1\) holds for all integers \(m \geq 5\). Therefore, we have \(n \leq 4\). Note that \(n\) is not equal to any of the integers \(1, 4\). Moreover, if \(n = 2\), then \(n \times 2^n + 1 = 9\), which is a perfect square. Finally, if \(n = 3\), then \(n \times 2^n + 1 = 25\), which is a perfect square. Thus, the only positive integer \(n\) such that \(n \times 2^n + 1\) is a perfect square is \(n = 2\) and \(n = 3\).