PUMaC 2012 · 加试 · 第 4 题

4. P ( X ) /Q ( X ) is not constant in X . Hint: Write down some linear equations and solve for the coefficients of P, Q using Problem 3.4! 9 ∑ ∑ n n i i Solution. We write P ( x ) = a x and Q ( x ) = b x and solve for the i i i =0 i =0 2 n + 2 unknown coefficients in a way that satisfies the above criteria. The third condition gives us that ( ) n ∑ i i − j i − j +1 ( a α + b α ) = 0 i i j i =0 ( ) i for 0 ≤ j < m where := 0 for j > i . By the last part of Question 3.1, we j have that ( ) d − 1 n ∑ ∑ i k α ( c a + c b ) = 0 i − j,k i i − j +1 ,k i j k =0 i = j r for 0 ≤ j < m and c < c where c > 1 depends only on α . This is true if r,k 0 0 k and only if the coefficients of α are zero for 0 ≤ k < d in each of the above m equations and hence we get dm linear equations in the 2 n + 2 unknowns a , b i i ( ) i n with integer coefficients c which are bounded in absolute value by (2 c ) . r,k 0 j Since (2 n + 2) > dm , by Question 3.4, we have the existence of solutions a , b i i bounded in absolute value by md/ (2 n +2 − md ) 1 /λ n/λ (2 n + 2) A < (2 n + 2) A ≤ (8 c ) 0 which is the bound required by picking c = 8 c . 1 0 We are left to show that the polynomials are both not identically zero and not constant multiples of each other. Assume without loss of generality that Q 6 = 0 but P = cQ for a constant c ∈ Q (possibly 0). By condition 3 , we have from 2.4 that R ( x ) := ( c + α ) Q ( x ) has at x = α a zero of multiplicity at least m since D ( R ( x ))( α ) = 0 for 0 ≤ j < m . Since c ∈ Q , we have that ( c + α ) 6 = 0 j − 1 and Q ( x ) = ( c + α ) R ( x ) has at x = α a zero of order at least m . This gives us deg( Q ) = n ≥ md > n + 1 since λ < 0 . 5 , a contradiction. 4.3 (10 points) Let d, n, λ, m, α, P, Q, c be as in the previous problem. Let u = p/q and v = r/s 1 be rational numbers in lowest terms such that q, s ≥ 2 and 1 1 | α − u | < and | α − v | < μ μ q s for some μ > 1 . Prove that for all 0 ≤ j < m , ( ) 1 1 n/λ | D ( P + vQ )( u ) | ≤ c + j 2 μ ( m − j ) μ q s for some c > 1 depending only on α . 2 Hint: Use the various facts about D and ‖ · ‖ from section 2 . k 10 Solution. Let F ( x, y ) = P ( x )+ yQ ( x ) . From the previous problem, we have that F ( x, a ) has a zero of multiplicity at least m at α and so F ( x, y ) = F ( x, α ) + m ( y − α ) Q ( x ) = ( x − α ) R ( x ) + ( y − α ) Q ( x ) , where R ∈ C [ x ] . This gives us m − j D F ( x, y ) = ( x − α ) S ( x ) + ( y − α ) D Q ( x ) by D ( f g ) = f D ( g ) + gD ( f ) and j j ( j ) D = j ! D where S ∈ C [ x ] . Now using results from section 2 and the fact that j 1 | u | , | v | < 1 , we get m − j | D ( F )( u, v ) | = | ( u − α ) S ( u ) + ( v − α ) D Q ( u ) | j j − μ ( m − j ) − μ ≤ q ( n + 1) ‖ S ‖ + s ( n + 1) ‖ D Q ‖ j n/λ m − j Now ‖ D ( Q ) ‖ ≤ (2 c ) and D F ( x, α ) = ( x − α ) S ( x ) . Thus, we get j 1 j from results of section 2 that n − j n/λ ‖ S ‖ < (deg S + 1)(2 / | α | ) ‖ D F ( x, α ) ‖ ≤ (16 c /α ) j 1 n n/λ since deg( S ) ≤ n < 2 , | α | < 1 and ‖ D P ‖ , ‖ D Q ‖ ≤ (2 c ) . Since 2( n +1) ≤ j j 1 n 4 , the desired estimate follows by choosing c = 64 c / | α | . 2 1 4.4 (12 points) Let d, n, λ, m, α, P, Q, u = p/q, v = r/s be as in the previous problem. Prove that D ( P + vQ )( u ) 6 = 0 h

for some h ∈ Z such that h ≤ 1 + ( c /λ ) n/ log q , where c > 0 depends only 3 3 on α . Note that log q = log q . e Hint: Recall part 4 of Problem 3.3. Solution. Observe that W := D ( P ) Q − D ( Q ) P 6 = 0 since P, Q are not propor- ( ) ∑ j j ( j ) ( i +1) ( j − i ) tional, by Question 2.5. We have D ( W ) = ( D ( P ) D ( Q ) − i =0 i ( j − i ) ( i +1) D ( P ) D ( Q )) by applying D ( f g ) = f D ( g ) + gD ( f ) iteratively. Let h be the minimum positive integer such that D ( P + vQ )( u ) 6 = 0 . We know h h exists since P + vQ 6 = 0 as a polynomial and so for 0 ≤ j < h , we have ( D ( P ) + vD ( Q ))( u ) = 0 . Eliminating v gives the equations ( D ( P ) D ( Q ) − j j j i − 1 ( j ) D ( P ) D ( Q ))( u ) = 0 for 0 ≤ i, j < h and thus D ( W ) = ( j !) D ( u ) = 0 for i j j 0 ≤ j < h − 1 and hence W has a zero of order at least h − 1 at x = u . We h − 1 know from part 4 of 3.3 that q ≤ ‖ W ‖ and 2 n/λ 2 n/λ ‖ W ‖ ≤ 2 n ‖ P Q ‖ ≤ 2 n (2 n + 1) c ≤ (4 c ) , 1 1 2 implying the desired result when c = log(4 c ) . 3 1 4.5 (22 points) Let 0 < < 1 / 2 . Prove that for all α ∈ Q of degree d ≥ 1 , ∣ ∣ ∣ ∣ p 1 ∣ ∣ α − < ∣ ∣ 1+ + d/ 2 q q 11 has only finitely many solutions for the rational p/q in lowest terms. Hint: Assume that there are infinitely many solutions. Let t be a an even integer such that t > 4 d/ − 2 and let μ = 1 + + d/ 2 . Given t , carefully select n, λ, m, P, Q, u = p/q, v = r/s as in the above problems ( u, v exist by the assumption of infinitely many solutions) and produce a contradiction between the results of Problems 4.3 and 4.4. Solution. By 3.1.2, it suffices to show this theorem for algebraic integers β , since for any α we can let β = kα , and if ∣ ∣ ∣ ∣ p c ∣ ∣ kα − > ∣ ∣ 1+ + d/ 2 q q p for any fraction 6 = kα , then q ∣ ∣ ∣ ∣ p c/k ∣ ∣ α − > ∣ ∣ 1+ + d/ 2 kq q p for any fraction 6 = α . Thus if we show c to exist in the first case, there are kq only finitely many solutions for p/q , by Lemma 4.1 in the solution for question 4.1. Using Question 4.1, we reduce to the case of d ≥ 3 and α is an algebraic ∣ ∣ ∣ ∣ p p 1+ + d/ 2 integer in I . Assume that α − < 1 /q for infinitely many ∈ Q . ∣ ∣ q q Choosing approximation: Fix even t such that λ = 2 / (2 + t ) < / 2 d and thus 1 0 < λ < and t ≥ 24 . Let n run through the arithmetic progression defined 12 by n = i ( t/ 2 + 1) d − 1 for i ∈ N and let m = (2 n + 2)(1 − λ ) /d = it . Pick 1 /λ 1 /λ 1 /λ c = max( c , c , c ) (from above) and set μ = 1 + + d/ 2 and δ = (1 + 1 2 3 p r 2 /d )(1 − λ ) − 1 . Select two rational approximations u = and v = from the q s infinitely many available such that ( p, q ) = ( r, s ) = 1 , 2 ≤ q < s and − μ