PUMaC 2020 · 代数（A 组） · 第 7 题
PUMaC 2020 — Algebra (Division A) — Problem 7
题目详情
- Suppose that p is the unique monic polynomial of minimal degree such that its coefficients are 2 π 4 π a rational numbers and one of its roots is sin + cos . If p (1) = , where a, b are relatively 7 7 b prime integers, find | a + b | .
解析
- Suppose that p is the unique monic polynomial of minimal degree such that its coefficients are 2 π 4 π a rational numbers and one of its roots is sin + cos . If p (1) = , where a, b are relatively 7 7 b prime integers, find | a + b | . Proposed by: Frank Lu Answer: 57 2 nπ 4 nπ We’ll first find the polynomial with roots that are sin + cos , where n goes from 1 to 7 7 6 and are integers. Then, we’ll show that this has minimal degree. Let this polynomial be q. 6 6 ∏ ∏ 2 nπ 4 nπ 2 2 nπ 2 nπ First, notice that ( x − sin − cos ) = ( x + 2 sin − sin − 1) . 7 7 7 7 n =1 n =1 6 6 ∏ ∏ 2 2 2 2 − x − 3 x − x − 3 x nπ 2 nπ However, notice also that q ( ) = ( + 2 sin − sin − 1) = ( x + 2 2 7 7 n =1 n =1 2 nπ x 2 nπ 2 sin + 1)( − + sin − 1) . Suppose that h is the monic polynomial with roots being the 7 2 7 2 nπ sin . Then, notice that this is equal to 64 h ( − ( x + 1) / 2) h ( x/ 2 + 1) . 7 We can explicitly find what h is, however. Notice that the equation giving that sin 7 θ = 0 , 7 5 3 2 4 using DeMoivre’s theorem, yields us the equation − sin θ + 21 sin θ cos θ − 35 sin θ cos θ + 7 5 2 3 2 6 2 7 sin θ cos θ = 0 , or that − sin θ + 21 sin θ (1 − sin θ ) − 35 sin θ (1 − sin θ ) + 7 sin θ (1 − 2 3 sin θ ) = 0 . 7 5 7 3 5 Expanding this out, we see that this is − sin θ + 21 sin θ − 21 sin θ − 35 sin θ + 70 sin θ − 7 3 5 7 7 5 35 sin θ +7 sin θ − 21 sin θ +21 sin θ − 7 sin θ = 0 . Simplifying, this is − 64 sin θ +112 sin θ − 3 3 56 sin θ + 7 sin θ = 0 . Notice that this has 7 roots, but one of these is just 0; this yields us 6 7 4 7 2 7 that, in fact, h ( x ) = x − x + x − . Furthermore, we see that this polynomial cannot be 4 8 64 factored further in the rationals; we can check this using Eisenstein’s criterion, for instance. From here, we will show that, in fact, q = p. Once we have this, we see that p (1) = q (1) = 2 − ( − 2) − 3 ∗ ( − 2) 1 − 7+7 · 2 − 7 − 7 7 q ( ) = 64 h (1 / 2) h (0) = 64( )( ) = − , which would yield our desired 2 64 64 64 answer of 57 . To show that p is q, we know that p has to divide q. But in fact, notice that q has to be at 2 π 2 least degree 3 , since p ( − 2 x + x + 1) is a polynomial where sin is a root, so is divisible 7 2 2 π 2 π 2 by a sixth degree polynomial h. But notice that − 2 x + x + 1 + 2 sin − sin − 1 = 7 7 2 π 2 π 2 ( x − sin )( − 2 x − 2 sin − 1) . However, notice that none of the other roots of p ( − 2 x + x + 1) 7 7 2 nπ 1 2 mπ are roots of h ; otherwise we have that − sin − = sin for some integers m, n, or that 7 2 7 1 2 mπ 2 nπ − = sin + sin . But we see that this doesn’t occur; indeed, notice that both of these 2 7 7 √ √ π π 6 − 2 1 sines can’t be negative (as sin > sin = > ), and if one is negative and one is 7 12 4 4 1 3 π π 2 π π positive, we require that either is sin − sin or sin − sin . 2 7 7 7 7 √ √ √ 3 π π π 3 2 − 6 None of these hold, though, as the second is 2 cos sin < 3 sin = , and the 14 14 12 2 2 π π π 3 π other is 2 cos sin . But if this is 1 / 2 , this means that 2 sin − 4 sin − 1 / 2 = 0 , which is 7 7 7 7 π not possible as we deduced that the polynomial of minimal degree is degree 6 for sin . 7 This forces us to have p = q, as desired.