Competition problem — Number theory

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Number theory Difficulty 6.1 National olympiad Prove it

13. Let $m>1,(a, m)=1$ . Prove: The binary linear congruence equation
$a x+y \equiv 0(\bmod m)$

must have a solution $x=x_{0}, y=y_{0}$ , satisfying $0<\left|x_{0}\right| \leqslant \sqrt{m}, 0<\left|y_{0}\right| \leqslant \sqrt{m}$ (Hint: Use the Pigeonhole Principle).

Solution

13. Consider the set $\{a x+y: 0 \leqslant x \leqslant[\sqrt{m}], 0 \leqslant y \leqslant[\sqrt{m}]\}$ , the number of its elements $=([\sqrt{m}]+1)^{2}>m$ .
Using the pigeonhole principle and $(a, m)=1$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.