PUMaC 2014 · 数论（A 组） · 第 8 题
PUMaC 2014 — Number Theory (Division A) — Problem 8
题目详情
- [ 8 ] Find all number sets ( a, b, c, d ) s.t. 1 < a ≤ b ≤ c ≤ d , a, b, c, d ∈ N , and a + b + c + d , 2 2 2 a + b + c + d , a + b + c + d and a + b + c + d are all square numbers. Sum the value of d across all solution set(s). 1
解析
- [ 8 ] Find all number sets ( a, b, c, d ) s.t. 1 < a ≤ b ≤ c ≤ d , a, b, c, d ∈ N , and a + b + c + d , 2 2 2 a + b + c + d , a + b + c + d and a + b + c + d are all square numbers. Sum the value of d across all solution set(s). Solutions: 2 2 2 2 2 2 We see that ( d + 2) > a + b + c + d > d Hence a + b + c + d = ( d + 1) = d + 2 d + 1. Hence we see that a + b + c = 2 d + 1. 3 7 2 2 2 2 We see that 2 d + 1 ≤ 3 c and thus d < c . Thus c < a + b + c + d < c + c < ( c + 2) . 2 2 2 2 2 We see that a + b + c + d = ( c + 1) = c + 2 c + 1. Hence a + b + d = 2 c + 1. With the previous equation, we see that d = c , and a + b = d + 1.Thus d + 1 ≤ 2 b and we have d < 2 b . 2 2 2 2 2 2 2 Thus b < a + b + c + d < b + b +4 b . Thus either a + b + c + d = ( b +1) or a + b + c + d = ( b +2) . 2 2 Case 1: a + b + c + d = ( b + 1) 2 2 We have a + c + d = 2 b + 1. Hence b = c = d . Thus a + b + c + d = ( b + 1) − 1 + a is a square. Hence a = 1 is the only solution, which is not in the range. 2 2 Case 2: a + b + c + d = ( b + 2) We have that a + c + d = 4 b + 4. We have from before c = d and a + b + c = 2 d + 1. Hence we can reduce the equations to a + 2 c = 4 b + 4 and a + b = c + 1. Hence a + 2( a + b − 1) = 4 b + 4, 2 3 5 2 2 which can be rewritten as a − 3 = b , and c = a − 4. Since ( a + 1) < a + b + c + d = 2 2 2 3 2 2 2 2 2 a + a − 3 + 5 a − 8 < ( a + 4) , we see that a + b + c + d = ( a + 2) or a + b + c + d = ( a + 3) . 2 2 2 Case 2a: a + b + c + d = ( a + 2) Thus b + c + d = 4 a + 4 which gives a = b . Hence b + c + d = 4 a + 4 and a + b + c = 2 d + 1 becomes 2 c = 3 a + 4 and 2 a = c + 1. Solving, we have 2(2 a − 1) = 4 a − 2 = 3 a + 4 which gives a = 6 and d = 11. 2 2 Case 2b: a + b + c + d = ( a + 3) Thus b + 2 c = 6 a + 9. We have a + 2 c = 4 b + 4. Thus we see that 5 b + 4 = 7 a + 9. Putting this 7 a 7 a into b + c + d = 6 a + 9 and a + b + c = 2 d + 1 we have 2 c = 6 a + 8 − and a + + 1 = c + 1. 5 5 Solving, we have that a = 40 , c = 96 and hence b = 57 and d = 96. Thus sum across all solution for d is 11 + 96 = 107 3