MOPSS, 13th December 2025
MOPSS Problems, Walkthroughs, Solutions from 13th December 2025. Notes for Mathematics Olympiad, IOQM, RMO, INMO. Problem set, Solutions, Questions, Answers, Hints, Walkthroughs, Discussions, Solutions in pdf.
(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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