MOPSS, 13th December 2025
(Japan Mathematical Olympiad 1992 P1) Let \(x, y\) be coprime positive integers with \(xy > 1\), and let \(n\) be an even positive integer. Prove that \(x^n + y^n\) is not divisible by \(x + y\).
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On the contrary, let us assume that \(x + y\) divides \(x^n + y^n\). Since \(x\) is congruent to \(-y\) modulo \(x + y\), and \(n\) is even, it follows that \(x^n\) is congruent to \(y^n\) modulo \(x + y\). This shows that \(x^n + y^n\) is congruent to \(2y^n\) modulo \(x + y\). Therefore, \(x + y\) divides \(2y^n\). Since \(x, y\) are coprime, it follows that \(x + y\) and \(y^n\) are coprime. Hence, \(x + y\) divides \(2\). Since \(x, y\) are positive integers with \(xy > 1\), it follows that \(x + y \geq 3\), which is a contradiction. Thus, \(x + y\) does not divide \(x^n + y^n\).
(Japan Mathematical Olympiad 1993 P2) For a positive integer \(n\), let \(d(n)\) denote its largest odd divisor. Define
\[\begin{align*} D(n) & = d(1) + d(2) + \dots + d(n),\\ T(n) & = 1 + 2 + \dots + n. \end{align*}\]Show that there exist infinitely many positive integers \(n\) such that \(3D(n) = 2T(n)\).
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- Check that \(n = 2, 6, 14, 30\) works.
- This may suggest that \(n = 2^m - 2\) works for all \(m \geq 2\).
- Does induction help?
Lecture notes for Math Olympiad (IOQM, RMO, INMO)
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