IOQM 2023 Questions, Solutions, Discussions

• 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 \(16\).

• 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 \(16\).

• 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 \(3^2 + 9^2 + 2^2 = 9 + 81 + 4 = 94\).