MOPSS, 18th October 2025

Let \(n\) be a positive integer. Note that

\(\begin{align*} 121^n - 25^n + 1900^n - (-4)^n & \equiv (-4)^n - 25^n + 25^n - (-4)^n \pmod{125}\\ & \equiv 0 \pmod{125} . \end{align*} \\)

Also note that

\(\begin{align*} 121^n - 25^n + 1900^n - (-4)^n & \equiv 9^n - 9^n + (-20)^n - (-20)^n \pmod{16}\\ & \equiv 0 \pmod{16} . \end{align*} \\)

Since \(125\) and \(16\) are coprime, it follows that \[ 121^n - 25^n + 1900^n - (-4)^n \] is divisible by \(2000\) for all positive integers \(n\).

Let \(n\) be an integer such that \(n^2 + 20n + 11\) is a perfect square. Then there exists a nonnegative integer \(m\) such that \[ n^2 + 20n + 11 = m^2. \] Note that \[ (n + 10)^2 - 89 = m^2 \] holds, which yields dm \[ (n + 10 - m)(n + 10 + m) = 89. \] Since \(89\) is prime and the integer \(m\) is nonnegative, it follows that \[ (n + 10 - m, n + 10 + m) \] is equal to one of \((1, 89)\) or \((-89, -1)\). In the first case, we have \[ n = 35, m = 44, \] and in the second case, we have \[ n = -55, m = 44. \]

Note that both \(n = 35\) and \(n = -55\) satisfy the condition that \(n^2 + 20n + 11\) is a perfect square. Therefore, the integers \(n\) for which \(n^2 + 20n + 11\) is a perfect square are \[ 35, -55. \]

Since \(p^2\) is the mean of \(u^2\) and \(v^2\), we have \[ u^2 + v^2 = 2p^2. \] Multiplying both sides by \(2\), we obtain \[ (u - v)^2 + (u + v)^2 = (2p)^2. \] This yields \[ (2p - u - v)(2p + u + v) = (u - v)^2. \] Let \(n\) be an odd positive integer dividing both \(2p - u - v\) and \(2p + u + v\). Then \(n\) divides their sum \(4p\) and their difference \(2(u + v)\). Since \(n\) is odd, it divides \(p\) and \(u + v\). As \(p\) is prime, we have \(n = 1\) or \(n = p\).

Let us consider the case that \(n = p\). Note that \(p\) divides \(u + v\) and \(u - v\). Since \(p\) is odd, it divides both \(u\) and \(v\). Since \(u\) and \(v\) are distinct positive integers, this implies that \[ u^2 + v^2 > 2p^2, \] which is a contradiction. This shows that \(n = 1\), and hence, \(2p - u - v\) and \(2p + u + v\) admit no common odd positive integer divisor other than \(1\). It follows that there are nonnegative integers \(a\) and \(b\), and relatively prime odd positive integers \(m\) and \(n\), such that \[ 2p - u - v = 2^a m, \quad 2p + u + v = 2^b n. \] Since \(2^{a + b} mn\) is a perfect square, we have that \(a + b\) is even and both \(m\) and \(n\) are perfect squares. This proves that \(2p - u - v\) is a square or twice a square.

Let \(n\) be a positive integer such that \(12n - 119\) and \(75n - 539\) are both perfect squares. Let \(a, b\) be nonnegative integers such that \[ 12n - 119 = a^2, \quad 75n - 539 = b^2. \] This yields

\(\begin{align*} (5a)^2 - (2b)^2 & = 5^2 \cdot 12 n - 5^2 \cdot 119 - 2^2 \cdot 75 n + 2^2 \cdot 539\\ & = 2^2 \cdot 420 - 119 \cdot (25 - 4)\\ & = 21 \cdot (80 - 119)\\ & = -21 \cdot 39\\ & = - 3^2 \cdot 7 \cdot 13, \end{align*} \\)

and this gives \[ (2b - 5a)(2b + 5a) = 3^2 \cdot 7 \cdot 13. \] Note that the factors on the left-hand side are both positive integers differing by a multiple of \(10\). Also note that \(3, 13\) are congruent to \(3 \pmod{10}\), and \(7\) is congruent to \(-3 \pmod{10}\). This shows that \((2b - 5a, 2b + 5a)\) is equal to one of the following pairs: \[ (3, 3 \cdot 7 \cdot 13), (7, 3^2 \cdot 13), (3 \cdot 7 \cdot 13, 3), (3^2 \cdot 13, 7) . \] Since \(a\) is nonnegative, we have \(2b + 5a \geq 2b - 5a\), so we only need to consider the first two cases. In the first case, we have \(a = 27\), and in the second case, we have \(a = 11\). Since \(12n - 119\) is equal to \(a^2\), and \(3\) does not divide \(119\), we have that \(a\) is not divisible by \(3\), and hence \(a\) is equal to \(11\). Note that \[ n = \frac{a^2 + 119}{12} = \frac{11^2 + 119}{12} = \frac{240}{12} = 20. \]

Observe that \[ 12 \cdot 20 - 119 = 121 = 11^2, \] and \[ 75 \cdot 20 - 539 = 1500 - 539 = 961 = 31^2 . \]

This proves that \(n = 20\) is the only positive integer such that both \(12n - 119\) and \(75n - 539\) are perfect squares.

Let \(x, y\) be integers satisfying the equation. Rearranging, we have \[ (x + y)^2 - xy = \left( \frac{x + y}{3} + 1 \right)^3 . \] Note that \(3\) divides \(x + y\). Write \(x + y = 3n\) for some integer \(n\). We obtain \[ 9n^2 - xy = (n + 1)^3, \] which yields \[ 9n^2 - x(3n - x) = n^3 + 3n^2 + 3n + 1, \] which simplifies to \[ x^2 - 3nx = n^3 - 6n^2 + 3n + 1. \] This gives \[ 4x^2 - 12nx + 9n^2 = 4n^3 - 15n^2 + 12n + 4, \] which shows that \[ (2x - 3n)^2 = (n - 2)(4n^2 - 7n - 2) , \] and hence, \[ (2x - 3n)^2 = (n - 2)^2(4n + 1) \] holds. This implies that \(4n + 1\) is a perfect square. Write \(4n + 1 = (2m + 1)^2\) for some integer \(m\). This gives \(n = m^2 + m\), and hence we obtain \[ (2x - 3n)^2 = (n - 2)^2 (2m + 1)^2, \] which implies that \(2x - 3n\) equals \(\pm (n - 2)(2m + 1)\). This shows that

\(\begin{align*} x & = \frac{3n + n(2m + 1)}{2} - 2m - 1\\ & = 2n + mn - 2m - 1\\ & = 2m^2 + 2m + m^3 + m^2 - 2m - 1\\ & = m^3 + 3m^2 - 1, \end{align*} \\)

holds or

\(\begin{align*} x & = \frac{3n - n(2m + 1)}{2} + 2m + 1\\ & = n - mn + 2m + 1\\ & = m^2 + m - m^3 - m^2 + 2m + 1\\ & = -m^3 + 3m + 1 \end{align*} \\)

holds. It follows that \((x, y)\) is equal to one of \[ (m^3 + 3m^2 - 1, -m^3 + 3m + 1) , \quad (-m^3 + 3m + 1, m^3 + 3m^2 - 1) . \]

Also note that if \(m, n\) are integers satisfying \(n = m^2 + m\), and \((x, y)\) is a pair of integers satisfying \(x + y = 3n\) and \[ (2x - 3n)^2 = (n - 2)^2 (2m + 1)^2, \] then it follows that \[ x^2 + xy + y^2 = \left( \frac{x + y}{3} + 1 \right)^3. \]

Consequently, the integer solutions to the equation are exactly the pairs of the form \[ (m^3 + 3m^2 - 1, -m^3 + 3m + 1) , \quad (-m^3 + 3m + 1, m^3 + 3m^2 - 1) , \] for some integer \(m\).

Let \(m\) be a positive integer and \(n\) be a positive odd integer such that the integers \(m, m + 1, m + 2\) and \(m + 3\) are divisible by \(n, n + 2, n + 4\) and \(n + 6\) respectively. Then \(m\) is at least as large as \[ \left( \frac{(n + 4)(n + 6) + 1}{2} + \left( \frac{\mathfrak{n} - 1}{2} \right) (n + 4)(n + 6) \right) (n + 2) - 1 , \] where

\(\begin{align*} \mathfrak{n} = \begin{cases} n & \text{if } 3 \text{ does not divide } n,\\ \frac{n}{3} & \text{if } 3 \text{ divides } n. \end{cases} \end{align*} \\)

Let \(k, r, s, t\) be positive integers such that \[ m = kn, \quad m + 1 = r(n + 2), \quad m + 2 = s(n + 4), \quad m + 3 = t(n + 6). \] Using that \(n\) divides \(m\), it follows that \(n\) divides \(2r - 1\).

Since \(n + 4\) divides \(m + 2\), it follows that \(n + 4\) divides \(2r - 1\). Finally, using that \(n + 6\) divides \(m + 3\), we obtain that \(n + 6\) divides \(2r - 1\). Hence, the integer \(2r - 1\) is divisible by each of \(n, n + 4, n + 6\). Since \(n + 4, n + 6\) are consecutive odd integers, they are coprime. Thus, \( (n + 4)(n + 6) \) divides \(2r - 1\). This shows that \[ r = \frac{(n + 4)(n + 6) + 1}{2} + \alpha (n + 4)(n + 6) \] for some non-negative integer \(\alpha\). Since \(n\) divides \(2r - 1\), it follows that \(n\) divides \(24 + 48 \alpha\). Using that \(n\) is odd, we obtain that \(n\) divides \(3(2\alpha + 1)\). Write

\(\begin{align*} \mathfrak{n} = \begin{cases} n & \text{if } 3 \text{ does not divide } n,\\ \frac{n}{3} & \text{if } 3 \text{ divides } n. \end{cases} \end{align*} \\)

Note that \(\mathfrak{n}\) is odd and \(\mathfrak{n}\) divides \(2\alpha + 1\). Thus, there exists a non-negative integer \(\beta\) such that \[ \alpha = \mathfrak{n} \beta + \frac{\mathfrak{n} - 1}{2}. \] Substituting this into the expression for \(r\), we obtain \[ r = \frac{(n + 4)(n + 6) + 1}{2} + \left( \mathfrak{n} \beta + \frac{\mathfrak{n} - 1}{2} \right) (n + 4)(n + 6) . \]

Let \(R\) denote the positive integer \[ \frac{(n + 4)(n + 6) + 1}{2} + \left( \frac{\mathfrak{n} - 1}{2} \right) (n + 4)(n + 6) , \] and let \(M\) denote the positive integer \[ R(n + 2) - 1. \] Then, the integers \(M, M + 1, M + 2, M + 3\) are divisible by \(n, n + 2, n + 4, n + 6\) respectively.

Note that

\(\begin{align*} 2R - 1 & \equiv 24 + 24 (\mathfrak{n} - 1) \pmod{n}\\ & \equiv 24 \mathfrak{n} \pmod{n}\\ & \equiv 0 \pmod{n}. \end{align*} \\)

This shows that \(n\) divides \(M\). Also note that \(n + 2\) divides \(M + 1\). Moreover,

\(\begin{align*} & M + 2\\ & = R(n + 2) + 1\\ & = \frac{(n + 2)(n + 4)(n + 6) + n + 4}{2} + \left( \frac{\mathfrak{n} - 1}{2} \right) (n + 2)(n + 4)(n + 6)\\ & = \frac{(n + 2)(n + 6) + 1}{2} (n + 4) + \left( \frac{\mathfrak{n} - 1}{2} \right) (n + 2)(n + 4)(n + 6) \end{align*} \\)

holds, which shows that \(n + 4\) divides \(M + 2\). Similarly, it follows that \[ M + 3 = \frac{(n + 2)(n + 4) + 1}{2} (n + 6) + \left( \frac{\mathfrak{n} - 1}{2} \right) (n + 2)(n + 4)(n + 6) , \] which shows that \(n + 6\) divides \(M + 3\). This completes the proof of the claim.

Combining the two claims, it follows that the smallest possible value of \(m\) is equal to \(M\), that is, the integer \[ \frac{(n + 2)(n + 4)(n + 6) + n}{2} + \left( \frac{\mathfrak{n} - 1}{2} \right) (n + 2)(n + 4)(n + 6) , \] which is equal to \[ \frac{1}{2} \left( n + \mathfrak{n} (n + 2)(n + 4)(n + 6) \right) . \]