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Iranian Mathematical Olympiad

Number theory Difficulty 5.7 AIME, harder Prove it Iran

We call a positive integer n lucky if it has at least two prime divisors and can be written in the form
n=p1α1+⋯+pkαk n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
where p1,…,pkp_1, \dots, p_k are distinct prime divisors of nn. (Note that nn may have other prime divisors than p1,…,pkp_1, \dots, p_k.) Prove that for every prime number pp, there exists a lucky number nn such that p∣np \mid n.

Solution

We claim that there exist nn with three prime factors. Choose a prime p1p_1. Using quadratic reciprocity and Dirichlet's theorem on arithmetic progressions, choose a prime p2p_2 such that
(p1p2)=(p2p1)=1 \left( \frac{p_1}{p_2} \right) = \left( \frac{p_2}{p_1} \right) = 1
and choose p3p_3 such that
(p1p3)=−1,(p2p3)=1 \left(\frac{p_1}{p_3}\right) = -1, \quad \left(\frac{p_2}{p_3}\right) = 1
Now, choose α1\alpha_1 such that its exponents modulo p2,p3p_2, p_3 are equal to p2−12,p3−12\frac{p_2-1}{2}, \frac{p_3-1}{2} respectively, and choose α2,α3\alpha_2, \alpha_3 similarly. Then n=p1α1+p2α2+p3α3n = p_1^{\alpha_1} + p_2^{\alpha_2} + p_3^{\alpha_3} satisfies the condition required in the problem. ■

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