HMMT 十一月 2011 · 团队赛 · 第 4 题
HMMT November 2011 — Team Round — Problem 4
题目详情
- [ 7 ] Determine the number of quadratic polynomials P ( x ) = p x + p x − p , where p , p , p are not 1 2 3 1 2 3 necessarily distinct (positive) prime numbers less than 50, whose roots are distinct rational numbers. C Coloring
解析
- [ 7 ] Determine the number of quadratic polynomials P ( x ) = p x + p x − p , where p , p , p are not 1 2 3 1 2 3 necessarily distinct (positive) prime numbers less than 50, whose roots are distinct rational numbers. Answer: 31 The existence of distinct rational roots means that the given quadratic splits into linear factors. Then, since p , p are both prime, we get that the following are the only possible factorizations: 1 3 • ( p x − p )( x + 1) ⇒ p = p − p 1 3 2 1 3 • ( p x + p )( x − 1) ⇒ p = − p + p 1 3 2 1 3 • ( p x − 1)( x + p ) ⇒ p = p p − 1 1 3 2 1 3 • ( p x + 1)( x − p ) ⇒ p = − p p + 1 1 3 2 1 3 In the first case, observe that since p + p = p , we have p > 2, so p is odd and exactly one of p , p 2 3 1 1 1 2 3 is equal to 2. Thus, we get a solutions for every pair of twin primes below 50, which we enumerate to be (3 , 5) , (5 , 7) , (11 , 13) , (17 , 19) , (29 , 31) , (41 , 43), giving 12 solutions in total. Similarly, the second case gives p + p = p , for another 12 solutions. 1 2 3 Team Round In the third case, if p , p are both odd, then p is even and thus equal to 2. However, this gives 1 3 2 p p = 3, which is impossible. Therefore, at least one of p , p is equal to 2. If p = 2, we get 1 3 1 3 1 p = 2 p − 1, which we find has 4 solutions: ( p , p ) = (3 , 2) , (5 , 3) , (13 , 7) , (37 , 19). Similarly, there are 2 3 2 3 four solutions with p = 2. However, we count the solution ( p , p , p ) = (2 , 3 , 2) twice, so we have a 3 1 2 3 total of 7 solutions in this case. Finally, in the last case p = − p p + 1 < − (2)(2) + 1 < 0, 2 1 3 so there are no solutions. Hence, we have a total of 12 + 12 + 7 = 31 solutions.