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PUMaC 2022 · 数论（B 组） · 第 3 题

PUMaC 2022 — Number Theory (Division B) — Problem 3

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

  1. Find the sum of all prime numbers p such that p divides 2 2 2 p + p +2 2 p − p +4 ( p + p + 20) + 4( p + p + 22) .
解析
  1. Find the sum of all prime numbers p such that p divides 2 2 2 p + p +2 2 p − p +4 ( p + p + 20) + 4( p + p + 22) . Proposed by Sunay Joshi Answer: 344 We claim that the primes are p = 2 , 61 , 281, yielding an answer of 2 + 61 + 281 = 344. First, 4 4 the expression is congruent to 20 + 4 · 22 modulo p by Fermat’s Little Theorem. Next, note 4 4 4 that by the Sophie-Germain Identity, we can rewrite the expression as 2 · (10 + 4 · 11 ) = 4 2 2 2 2 6 2 · (10 + 2 · 11 − 2 · 10 · 11)(10 + 2 · 11 + 2 · 10 · 11), which equals 2 · 61 · 281. Since p divides this product, p must be among { 2 , 61 , 281 } , and the result follows.