HMMT 二月 2006 · 代数 · 第 7 题

HMMT February 2006 — Algebra — Problem 7

Let 4 3 2 f ( x ) = x − 6 x + 26 x − 46 x + 65 . Let the roots of f ( x ) be a + ib for k = 1 , 2 , 3 , 4. Given that the a , b are all integers, find k k k k | b | + | b | + | b | + | b | . 1 2 3 4 4 2

解析

Let 4 3 2 f ( x ) = x − 6 x + 26 x − 46 x + 65 . Let the roots of f ( x ) be a + ib for k = 1 , 2 , 3 , 4. Given that the a , b are all integers, k k k k find | b | + | b | + | b | + | b | . 1 2 3 4 Answer: 10 Solution: The roots of f ( x ) must come in complex-conjugate pairs. We can then say that a = a and b = − b ; a = a and b = − b . The constant term of f ( x ) is 1 2 1 2 3 4 3 4 2 2 2 2 the product of these, so 5 · 13 = ( a + b )( a + b ). Since a and b are integers 1 1 3 3 k k for all k , and it is simple to check that 1 and i are not roots of f ( x ), we must have 2 2 2 2 a + b = 5 and a + b = 13. The only possible ways to write these sums with 1 1 3 3 2 2 2 2 positive integers is 1 + 2 = 5 and 2 + 3 = 13, so the values of a and b up to sign 1 1 3 are 1 and 2; and a and b up to sign are 2 and 3. From the x coefficient of f ( x ), we 3 3 get that a + a + a + a = 6, so a + a = 3. From the limits we already have, this 1 2 3 4 1 3 tells us that a = 1 and a = 2. Therefore b , b = ± 2 and b , b = ± 3, so the required 1 3 1 2 3 4 sum is 2 + 2 + 3 + 3 = 10. 4 2