PUMaC 2011 · 数论（A 组） · 第 8 题
PUMaC 2011 — Number Theory (Division A) — Problem 8
题目详情
- [ 8 ] Calculate the sum of the coordinates of all pairs of positive integers ( n, k ) such that ( ) 2 n k ∑ ∑ 3 2 2 k ≡ 0 , 3 (mod 4), n > k , and i = (96 · 3 − 1) i + 48 . i =1 i = k +1 1
解析
- Note that ( ) 2 m m ∑ ∑ 3 i = i i =1 i =1 for all positive integers m . Therefore, ( ) ( ) n 2 2 ∑ n ( n + 1) k ( k + 1) 3 i = − . 2 2 i = k +1 4 The given equation therefore is the same as ( ) ( ) 2 2 n ( n + 1) k ( k + 1) 2 2 = (96 (3)) + 48 . 2 2 n ( n +1) 2 Since 48 divides the right side of this equation, 48 | , so we can write this equation as 2 ( ) 2 n ( n + 1) 2 − 3 ( k ( k + 1)) = 1 . 96 √ √ m m n ( n +1) (2+ 3) +(2 − 3) This is Pell’s equation. Therefore, can be written as for any positive 96 2 √ √ m m (2+ 3) − (2 − 3) √ integer m . Furthermore, k ( k + 1) = . Since k ≡ 0 , 3 (mod 4), 4 | k ( k + 1) and 2 3 √ √ m m (2+ 3) − (2 − 3) √ 8 | . Note that 3 √ √ m m (2 + 3) − (2 − 3) m b c 2 √ ≡ 2(3) (mod 8) 3 for odd m (since all terms besides the last term in the expansion are divisible by 8). This contradicts the fact that k ( k + 1) is a multiple of 4, so m is even. For even m , let m = 2 p for some integer p . Then √ √ m m n ( n + 1) (2 + 3) + (2 − 3) = 96 2 is equivalent to ( ) 2 √ √ √ √ √ 2 2 p 2 p p p 4 n + 4 n = 192((2 + 3) + (2 − 3) ) = 64 3((2 + 3) − (2 − 3) ) + 384 . √ √ √ p p 2 Note that 3((2+ 3) − (2 − 3) ) is a positive integer, so WLOG call it x . Then (2 n +1) = 2 64 x + 385, so from difference of squares (2 n + 1 − 8 x )(2 n + 1 + 8 x ) = 5 · 7 · 11. The only positive √ √ √ p p integer solutions to this equation are ( n, x ) = (96 , 24) and (15 , 3). 3 6 = 3((2+ 3) − (2 − 3) ) for any p , but p = 2 yields the x = 24 case. Therefore, the only possible solution occurs when n = 96. Plugging this into the original equation shows that k = 7 so the answer is 96+7 = 103 . 5