PUMaC 2010 · 代数（A 组） · 第 26 题
PUMaC 2010 — Algebra (Division A) — Problem 26
题目详情
- Find an integer n such that p ( n ) = n + 82. 1
解析
- Assume that p ( n ) = n + 82 for some integer n . Find n . Solution: 28. Since p ( n ) − n − 82 = 0, the polynomial p ( x ) − x − 82 must factor to ( x − n ) q ( x ), where q ( x ) is another polynomial. The polynomial q will have integer coefficients, because 2 j p ( x ) − x − 82 = p ( x ) − p ( n ) + n − x , so if we let p ( x ) = a + a x + a x + · · · + a x , we get 0 1 2 j 2 2 j j p ( x ) − x − 82 = ( a − 1)( x − n ) + a ( x − n ) + · · · + a ( x − n ) . 1 2 j i i Dividing through by x − n clearly leaves a polynomial with integer coefficients, since x − n is always divisible by x − n . In particular, therefore, q (15) and q (35) are integers, so plugging in 15 and 35 we get that 15 − n is divisible by p (15) − 15 − 82 = − 91 and 35 − n is divisible by p (35) − 35 − 82 = − 91. Since the factors of 91 are just ± 1, ± 7, and ± 13, we must have either 15 − n = − 13 = ⇒ n = 28 or 15 − n = − 7 = ⇒ n = 22. The latter case is ruled out because p (22) = 116 6 = 22 + 82. 3