PUMaC 2009 · 数论（B 组） · 第 5 题

PUMaC 2009 — Number Theory (Division B) — Problem 5

Find the sum of all prime numbers p which satisfy 4 4 4 p = a + b + c − 3 for some primes (not necessarily distinct) a , b and c . 3 3

解析

Find the sum of all prime numbers p which satisfy 4 4 4 p = a + b + c − 3 for some primes (not necessarily distinct) a , b and c . Solution. 719. If a , b and c are all odd, then the right hand side is even (and it’s greater than 2, which can be easily checked), and so this forces p to be an even number greater than 2, a contradiction. So exactly one or three of a , b and c is 2. Again, if all three are 2, then p = 45, which is not a prime, hence inadmissible. So exactly one of a , b and c is 2, say a . Then we 4 4 have p = b + c + 15. Now, if none of b and c is 3, then they are each of the form 6 k ± 1 for ′ some integer k . Then, their fourth powers are of the form 6 k + 1, and hence, adding them together, the right hand side becomes divisible by 3, which is inadmissible. So, one of b and c must be 3 (they cannot both be 3, because then p = 175, not a prime. So suppose b is 3. Then 4 we get p = c + 94. The last deduction is as follows: if c 6 = 5, then c must end in 1, 3, 7 or 9. The fourth power of each number of this form ends in the digit 1. Then, adding that to 94, we will get a number divisible by 5, a contradiction. So c must be 5. We have to finally check that p = 719 is indeed a prime. This is checked easily, and hence we get our unique solution. 3 3