HMMT 十一月 2017 · 团队赛 · 第 8 题

HMMT November 2017 — Team Round — Problem 8

[ 55 ] Find the number of quadruples ( a, b, c, d ) of integers with absolute value at most 5 such that 2 2 2 2 2 2 2 ( a + b + c + d ) = ( a + b + c + d )( a − b + c − d )(( a − c ) + ( b − d ) ) .

解析

[ 55 ] Find the number of quadruples ( a, b, c, d ) of integers with absolute value at most 5 such that 2 2 2 2 2 2 2 ( a + b + c + d ) = ( a + b + c + d )( a − b + c − d )(( a − c ) + ( b − d ) ) . Proposed by: Mehtaab Sawhney Answer: 49 Let x = a + c , y = a − c , w = b + d , and z = b − d . Then 2 2 2 2 2 2 2 2 2 ( w + x + y + z ) = 4( x − w )( y + z ) 2 2 2 2 and since | x + w | ≥ | x − w | it follows that w = 0 or y = z = 0. Now y = z = 0 implies 2 2 2 a = b = c = d = 0. Now w = 0 gives b = − d . Then for equality to hold x = y + z . This is equivalent 2 to ac = b , which includes the previous case. It suffices to count the number of triples ( a, b, c ) that satisfy the equation. When b = 0, either a or c is zero, which gives 11 + 11 − 1 = 21 triples. When b = ± 1, we have | a | = | c | = 1 and a, c have the same sign, for 2 · 2 = 4 triples. When b = ± 2, we have ( a, c ) = (1 , 4) , (2 , 2) , (4 , 1) or their negatives, for 2 · 6 = 12 triples. When b = ± 3 , ± 4 , ± 5, we have | a | = | b | = | c | and a, c have the same sign, for 6 · 2 = 12 triples. So in total there are 21 + 4 + 12 + 12 = 49 solutions.