MOPSS, 9th August 2025

• Note that \(x \neq 1\) and \(x - 1\) divides \(2\). It follows that \(x = 2\) or \(x = 3\).
• If \(x = 2\), then \(y^3 = 23 + 2^4 + 1 = 40\), which is impossible since \(y\) is an integer.
• If \(x = 3\), then \(y^3 = 23 + \frac 12 (3^4 + 1) = 23 + 41 = 64\), which gives \(y = 4\).
• It follows that \(x + y = 7\).

• Let \(r\geq 3\) be an integer and consider a regular \(r\)-gon. Note that \(360/r\) is even if and only if \(r\) is equal to one of

  \[ 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 24, 36, 45, 60, 120, 360 . \]

  In other words, there are precisely \(16\) positive integers \(r \geq 3\) such that dividing \(360\) by \(r\) yields an even number.

• It follows that the number of elements of \(X\) is equal to \(16\).

• Suppose \((a, b, c)\) is a pair satisfying the above conditions.
• Since \(ab\) is a prime, it follows that one of \(a\), \(b\) is equal to \(1\), and the remaining one is a prime.
• If \(a = 1\), then \(b, c\) are distinct primes.
• If \(b = 1\), then \(c\) is the product of two distinct primes which are not equal to \(a\).
• This shows that \((a, b, c)\) is equal to one of

\(\begin{align*} & (1, 2, 3), (1, 2, 5), (1, 2, 7), (1, 2, 11), (1, 2, 13), \\ & (1, 3, 2), (1, 3, 5), (1, 3, 7), \\ & (1, 5, 2), (1, 5, 3), \\ & (1, 7, 2), (1, 7, 3), \\ & (1, 11, 2), \\ & (1, 13, 2), \\ & (2, 1, 3 \cdot 5), (3, 1, 2 \cdot 5), (5, 1, 2 \cdot 3). \end{align*} \\)

• Hence, the number of triples satisfying the above conditions is at most \(17\).
• Since any of the above triples does satisfy the given conditions, it follows that the number of triples satisfying the given conditions is equal to \(17\).

• Compute the first few terms of the sequence to note that
  \(\begin{align*} a_0 & = 1, \\ a_1 & = - 4,\\ a_2 & = 9, \\ a_3 & = -8, \\ a_4 & = - 31, \end{align*}\)
  which gives

\(\begin{align*} a_1^2 - a_0a_2 & = 16 - 9 \\ & = 7,\\ a_2^2 - a_1a_3 & = 81 - 32 \\ & = 49,\\ a_3^2 - a_2a_4 & = 64 + 279 \\ & = 343. \end{align*} \\)

This may suggest that \[ a_n^2 - a_{n-1}a_{n+1} = 7^n \]

holds for any integer \(n\geq 1\).

• Indeed, it is clear for \(n = 1\). Assuming that \[ a_n^2 - a_{n-1}a_{n+1} = 7^n \] holds for some integer \(n\geq 1\), note that
  \(\begin{align*} & a_{n+1}^2 - a_{n} a_{n+2} \\ & = a_{n+1}^2 - a_n (-4a_{n+1} - 7a_n)\\ & = a_{n+1}^2 + 4a_n a_{n+1} + 7 a_n^2 \\ & = a_{n+1}^2 + 4a_n a_{n+1} + 7 a_{n-1}a_{n+1} + 7^{n+1} \\ & = a_{n+1} (a_{n+1} + 4a_n + 7a_{n-1}) + 7^{n+1}\\ & = 7^{n+1}. \end{align*}\)
  This proves that \[ a_n^2 - a_{n-1}a_{n+1} = 7^n \] holds for any integer \(n\geq 1\).
• Consequently, \[ a_{50}^2 - a_{49}a_{51} = 7^{50}, \] whose number of positive integer divisors is equal to \(51\).

• Let \(m,n\) be integers satisfying \(m^2 = 4n^2 - 5n + 16\), or equivalently, \[ 4n^2 - 5n + (16 - m^2) = 0. \]
• Since \(n\) is an integer, the discriminant of the quadratic polynomial \(4n^2 - 5n + (16 - m^2)\) in \(n\) is a perfect square, that is, there exists an integer \(\ell\) such that \[ 25 - 16 (16 - m^2) = \ell^2, \] or equivalently, \(16m^2 - \ell^2 = 231\), which gives \[(4m - \ell)(4m + \ell) = 3 \cdot 7 \cdot 11.\]
• Note that \(4m-\ell, 4m + \ell\) are odd and hence congruent to \(1, -1\) modulo \(4\) (in some order). Also note that \(4m - \ell < 4m + \ell\) holds.
• This shows that \((4m-\ell, 4m + \ell)\) is equal to one of \[ (1, 3\cdot 7 \cdot 11), (3, 77), (7, 33), (11, 21), \] and consequently, \(8m = 232, 80, 40, 32\), which yields \(m = 29, 10, 5, 4\).
• If \(m = 29\), then \(4n^2 - 5n - 825 = 0\), which gives \(n = 15\).
• If \(m = 10\), then \(4n^2 - 5n - 84 = 0\) holds, which gives \(n = - 4\).
• If \(m = 5\), then \(4n^2 - 5n - 9 = 0\) holds, implying \(n = -1\).
• If \(m = 4\), then \(n = 0\).
• Conclude that the maximum possible value of \(\lvert m - n \rvert\) is equal to \(14\).

• Note that if \(p_1 + p_2 + p_3 \neq 0\), then using the condition that \(p_1, p_2, p_3\) are real, it follows that \(p_1 = p_2 = p_3\).
• Note that

\(\begin{align*} p_1 & = 1 + a + b + c, \\ p_2 & = 8 + 4a + 2b + c, \\ p_3 & = 27 + 9a + 3b + c, \end{align*} \\)

which gives that \[ p_1 + p_2 + p_3 = 36 + 14a + 6b + 3c. \] Since \(c\) is an odd integer, it follows that¹ \(p_1 + p_2 + p_3 \neq 0\). Consequently, \(p_1 = p_2 = p_3\).

• This gives \[ 3a + b = -7, 5a + b = -19, \] and hence, \(a = -6, b = 11\).
• This gives
  \(\begin{align*} p_2 + 2p_1 - 3p_0 & = 10 + 6a + 4b + 3c - 3c \\ & = 18. \end{align*}\)

• The sides of the hexagon has been colored red.
• Note that if \(A_iA_jA_k\) is such that some two of its vertices are consecutive, then it has at least one red side.
• If \(A_iA_jA_k\) is such that no two of its vertices are consecutive, then any two of its vertices are exactly one vertex apart. Note that there precisely two such triangles, namely, \(A_1A_3A_5\) and \(A_2A_4A_6\).
• Observe that if each of these two triangles have at least one red side, then the required condition is satisfied.
• Moreover, if the required condition is satisfied, then each of these two triangles has at least one red side.
• It follows that \(N\) is equal to the number of colorings of the diagonals of the hexagon using red and blue such that each of the triangles \(A_1A_2A_3\) and \(A_2A_4A_6\) have at least one red side.

• This shows that \[ N = (2^3 - 1)(2^3 - 1)(2^{\binom{6}{2} - 6 - (3 + 3)}) = 7 \cdot 7 \cdot 8 = 392 . \]

• The sum of the squares of the digits of \(N\) is equal to \(3^2 + 9^2 + 2^2 = 9 + 81 + 4 = 94\).

Decompose the set of vertices of the \(20\)-gon using the vertices of the pentagons as in the image below.

• Show that the product of the entries in some subgrid is divisible by \(32\).
• Conclude that the product of all the \(16\) entries is divisible by \(16 \times 16 \times 16 \times 32\).
• Is the product of the integers \(17, 18, \dots, 32\) divisible by \(16 \times 16 \times 16 \times 32\)?

• Assume that the conclusion is false, and by interchanging the colors (if necessary), assume that there are more red points than the blue ones.
• Show that one of the four corners is red, and assume without loss of generality that the point \(E\) is red.
• Considering the number of red points in each of the sets \(\{A, A'\}\), \(\{B, B'\}\), \(\{C, C'\}\), \(\{D, D'\}\), prove that these four sets together contain at most \(4\) red points, and conclude that the green square contains at least \(8\) red points.
• Consider the blue squares in the green square, to show that the green square contains exactly \(8\) red points.
• Conclude that there are exactly \(13\) red points, and each of the sets \(\{A, A'\}\), \(\{B, B'\}\), \(\{C, C'\}\), \(\{D, D'\}\) contains exactly one red point, and the green square contains exactly \(8\) red points.
• For each point on the dashed diagonal, consider the square having this point and \(E\) as its vertices, to show that the points on the dashed diagonal are blue.
• Note that each point within the green square lying outside the dashed diagonal, is a vertex of a square having two of its remaining vertices on the dashed diagonal. Conclude that each such point is red.
• Obtain a contradiction to conclude that the assmption is false.

• Show that if the small square (as in Fig. 1) does not contain a blue point, then we are done.
• It remains to consider the case when the small square (as in Fig. 1) contains at least one blue point.
• Rotating the configuration about the center of the small square (if necessary), assume that the top-left vertex of the small square (as in Fig. 1) is blue.

• Prove that the gray square contains at most three blue points.

  • Consider the case when at least two blue points lie on the bigger dashed circle. Show that the smaller dashed circle does not contain any blue point, in this case. Hence, the gray square contains at most three blue points.

  • Similarly, if the smaller dashed circle contains at least two blue points, then the gray square (as in Fig. 2) contains at most three blue points.

• It suffices to consider that each one of the red, purple, and green \(L\)-shapes, has at most one end-point which is blue (otherwise, we are done).
• Use the above to show that the bottom-right point (as in Fig. 2) is blue.
• Consider the point at the bottom-right corner, and the center of the gray square, and a blue end-point of an \(L\)-shape, to show that these three points form the vertices of an isosceles triangle having the required properties.