IOQM 2026 class 1 - Parity and power of integers
Q1:
Find all nonnegative integers \( n \) such that there are integers \( a \) and \( b \) with the property
\[
n^2 = a + b \quad \text{and} \quad n^3 = a^2 + b^2.
\]
Solution:
Using QM-AM inequality, \[ \frac{a + b}{2} \le \sqrt{\frac{a^2 + b^2}{2}}. \] Substitute \( a + b = n^2 \) and \( a^2 + b^2 = n^3 \): \[ \frac{n^2}{2} \le \sqrt{\frac{n^3}{2}}. \] Squaring both sides: \[ \frac{n^4}{4} \le \frac{n^3}{2}. \] Simplify: \[ \frac{n}{2} \le 1. \] Thus, \[ n \le 2. \] So \( n = 0, 1, 2 \).
Q2:
In each box of a \(15 \times 15\) square table one of the numbers \(1,2,3,\dots,15\) is written. Boxes which are symmetric with respect to one of the main diagonals contain equal numbers, and no row or column contains two copies of the same number. Show that no two of the numbers along the main diagonal are the same.
Solution:
Sample 3x3 matrix:
[1,2,3]
[3,1,2]
[2,3,1]
Consider a smaller \(3 \times 3\) matrix to understand the structure. Let the matrix be symmetric about a main diagonal. Then for all \(i,j\) with \(i \ne j\), \[ A[i,j] = A[j,i]. \] Now fix a number and count how many times it appears in the matrix. Let it appear \(d\) times on the diagonal and \(2p\) times off the diagonal (since off-diagonal entries come in symmetric pairs). Then the total number of occurrences is \[ d + 2p. \] Since each row contains exactly one copy of each number, the number must appear exactly once in each row. For a \(3 \times 3\) matrix, this gives \[ d + 2p = 3. \] Thus, \[ d = 3 - 2p, \] so \(d\) is odd. Hence \(d = 1\) or \(3\). But if \(d = 3\), then the same number appears three times on the diagonal, which would force repetition in rows or columns, a contradiction. Therefore, \[ d = 1. \] Thus each number appears exactly once on the diagonal. The same argument applies to a \(15 \times 15\) matrix. Since \(15\) is odd, each number appears an odd number of times on the diagonal, hence exactly once. Therefore, all diagonal entries are distinct.
Q3.
Let \( n \) be a positive integer and let \( \varepsilon_1, \varepsilon_2, \dots, \varepsilon_n \in \{-1,1\} \) be such that \[ \varepsilon_1 \varepsilon_2 + \varepsilon_2 \varepsilon_3 + \cdots + \varepsilon_n \varepsilon_1 = 0. \] Prove that \( n \) is divisible by \( 4 \).
Solution:
\( \varepsilon_1 \varepsilon_2 + \varepsilon_2 \varepsilon_3 + \cdots - \varepsilon_n \varepsilon_1 = 0 \) (1) \(\Rightarrow\) half are \(-1\) and half \(= +1\) \(\Rightarrow\) \( n = 2k \) (even) multiply them all \(\Rightarrow\) \[ \varepsilon_1^2 \varepsilon_2^2 \cdots \varepsilon_n^2 = 1 \quad (2) \] \[ = \varepsilon_1 \varepsilon_2 \cdot \varepsilon_2 \varepsilon_3 \cdots \varepsilon_n \varepsilon_1 \] \(\therefore\) all are squared but earlier we said half are \(-1\) \(\Rightarrow\) occurs even number of times In (1), half the numbers are \(-1\) but all multiplied together give result is \(+1\) \(\Rightarrow\) they are also even so (1) has even \(-\)'s and even \(+\)'s \(\Rightarrow\) \[ n = 2p + 2p = 4p \]
Q4. Will they rotate together?
Solution:
Anti one gear counterclockwise \(\Rightarrow\) neighbor clockwise If odd gears connected together \(\Rightarrow\) A C C For eg: 3 \(\Rightarrow\) contradiction Neighbor can't be same Only even will rotate.
Using QM-AM inequality, \[ \frac{a + b}{2} \le \sqrt{\frac{a^2 + b^2}{2}}. \] Substitute \( a + b = n^2 \) and \( a^2 + b^2 = n^3 \): \[ \frac{n^2}{2} \le \sqrt{\frac{n^3}{2}}. \] Squaring both sides: \[ \frac{n^4}{4} \le \frac{n^3}{2}. \] Simplify: \[ \frac{n}{2} \le 1. \] Thus, \[ n \le 2. \] So \( n = 0, 1, 2 \).
Q2:
In each box of a \(15 \times 15\) square table one of the numbers \(1,2,3,\dots,15\) is written. Boxes which are symmetric with respect to one of the main diagonals contain equal numbers, and no row or column contains two copies of the same number. Show that no two of the numbers along the main diagonal are the same.
Solution:
Sample 3x3 matrix:
[1,2,3]
[3,1,2]
[2,3,1]
Consider a smaller \(3 \times 3\) matrix to understand the structure. Let the matrix be symmetric about a main diagonal. Then for all \(i,j\) with \(i \ne j\), \[ A[i,j] = A[j,i]. \] Now fix a number and count how many times it appears in the matrix. Let it appear \(d\) times on the diagonal and \(2p\) times off the diagonal (since off-diagonal entries come in symmetric pairs). Then the total number of occurrences is \[ d + 2p. \] Since each row contains exactly one copy of each number, the number must appear exactly once in each row. For a \(3 \times 3\) matrix, this gives \[ d + 2p = 3. \] Thus, \[ d = 3 - 2p, \] so \(d\) is odd. Hence \(d = 1\) or \(3\). But if \(d = 3\), then the same number appears three times on the diagonal, which would force repetition in rows or columns, a contradiction. Therefore, \[ d = 1. \] Thus each number appears exactly once on the diagonal. The same argument applies to a \(15 \times 15\) matrix. Since \(15\) is odd, each number appears an odd number of times on the diagonal, hence exactly once. Therefore, all diagonal entries are distinct.
Q3.
Let \( n \) be a positive integer and let \( \varepsilon_1, \varepsilon_2, \dots, \varepsilon_n \in \{-1,1\} \) be such that \[ \varepsilon_1 \varepsilon_2 + \varepsilon_2 \varepsilon_3 + \cdots + \varepsilon_n \varepsilon_1 = 0. \] Prove that \( n \) is divisible by \( 4 \).
Solution:
\( \varepsilon_1 \varepsilon_2 + \varepsilon_2 \varepsilon_3 + \cdots - \varepsilon_n \varepsilon_1 = 0 \) (1) \(\Rightarrow\) half are \(-1\) and half \(= +1\) \(\Rightarrow\) \( n = 2k \) (even) multiply them all \(\Rightarrow\) \[ \varepsilon_1^2 \varepsilon_2^2 \cdots \varepsilon_n^2 = 1 \quad (2) \] \[ = \varepsilon_1 \varepsilon_2 \cdot \varepsilon_2 \varepsilon_3 \cdots \varepsilon_n \varepsilon_1 \] \(\therefore\) all are squared but earlier we said half are \(-1\) \(\Rightarrow\) occurs even number of times In (1), half the numbers are \(-1\) but all multiplied together give result is \(+1\) \(\Rightarrow\) they are also even so (1) has even \(-\)'s and even \(+\)'s \(\Rightarrow\) \[ n = 2p + 2p = 4p \]
Q4. Will they rotate together?
Solution:
Anti one gear counterclockwise \(\Rightarrow\) neighbor clockwise If odd gears connected together \(\Rightarrow\) A C C For eg: 3 \(\Rightarrow\) contradiction Neighbor can't be same Only even will rotate.
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