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Number theory Difficulty 5.4 AIME, harder Find the answer

2 Let p1<p2<p3<p4<p5p_{1}<p_{2}<p_{3}<p_{4}<p_{5} be 5 prime numbers, and p1,p2,p3,p4,p5p_{1}, p_{2}, p_{3}, p_{4}, p_{5} form an arithmetic sequence. Find the minimum value of p5p_{5}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

2. Let dd be the common difference, then p1,p1+d,p1+2d,p1+3d,p1+4dp_{1}, p_{1}+d, p_{1}+2 d, p_{1}+3 d, p_{1}+4 d are all primes. If 2d2 \nmid d, i.e., dd is odd, then one of p1+dp_{1}+d and p1+2dp_{1}+2 d is even, and it is not a prime. If 3d3 \nmid d, then one of p1+d,p1+2d,p1+3dp_{1}+d, p_{1}+2 d, p_{1}+3 d is a multiple of 3 (they form a complete residue system modulo 3), which is a contradiction.

If 5d5 \nmid d, then one of p1,p1+d,,p1+4dp_{1}, p_{1}+d, \cdots, p_{1}+4 d is a multiple of 5, which can only be p1=p_{1}= 5, in this case the common difference dd is a multiple of 6.

And 5,11,17,23,295,11,17,23,29 is a sequence of 5 primes in arithmetic progression, so, p5p_{5} is at least 29.

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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.