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Mongolian Mathematical Olympiad

Number theory Difficulty 5.4 AIME, harder Prove it Mongolia

Let m∈Nm \in \mathbb{N}. m2<a,b<m2+mm^2 < a, b < m^2 + m and a≠ba \neq b. Find all the natural cc, such that c∣abc \mid ab, m2<c<m2+mm^2 < c < m^2 + m.

(proposed by D. Ganzorig)

Solution

Let dd be a number such that d∣abd \mid ab and d∈(m2,m2+m)d \in (m^2, m^2 + m). Then d∣(a−d)(b−d)d \mid (a-d)(b-d) and ∣a−d∣<m|a-d| < m, ∣b−d∣<m|b-d| < m. It follows that ∣(a−d)(b−d)∣<m2<d|(a-d)(b-d)| < m^2 < d. Hence
(a−d)(b−d)=0. (a-d)(b-d) = 0.
We have d=a∨bd = a \lor b.

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