By von zur Gathen J., Gerhard J.
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Extra resources for Modern computer algebra, Solutions for selected exercises
R Q j 1 ¡¡¡ Q1 , and a1 aj 4. r j 1 rj n j 1 nj n0 . a j 1 xn0 2d a j xn0 2d deg r j 1 deg r j Modern Computer Algebra, J OACHIM VON ZUR ·R ´ d µ, r0 a0 xn0 2d r1 a1 xn0 2d G ATHEN and J ÜRGEN G ERHARD, version 14 September 2003 46 Solutions to Chapter 11 0 or k 5. if r j 6. q j r j·1 7. d £ n0 n j then return j 1, R, and r j 1 rj r j 1 quo r j , j 1 lc´r j 1 rem r j µ, ´r j 1 rem r j µ i 11, n j 1 deg r j 1 k ´n0 n j µ r j 2d £ , a£j 1 r j 1 ´2d £ ´n j n j · · · · 8. a£j · ·1 µµ · call the algorithm recursively with input a£j a£j·1 and d £ , giving h j ´d £ µ, 9.
27 (i) follows by induction on n, (iii) by induction on r, (ii) is a special case of (iii), and (iv) follows from (ii) by dividing by f1 ¡¡¡ fr . 20, exists with a rational function ¾ R´yµ whose denominator is not divisible by y g. 22 works if ³ and ³¼ are defined at g0 modulo p, ³´g0 µ 0 mod p, and ³¼ ´g0 µ is invertible modulo p. For ³ f y 1, the Newton formula is gi gi 1 ´ f gi 1 1µ f 1 f , and it does not lead to an algorithm for computing 1 f . 30 The roots of ³ modulo 5 are 2 and 3, and the only root of ³ in 2 4 is 18.
In the remaining case where 2 p ¹ a, we have always S p ´aµ 1, for if g is a zero of ³ modulo p, then so is g g mod p. Each of the two cases occurs: for example, we have S3 ´2µ 0 and S3 ´1µ 2. (ii) The claim is clear if S p ´aµ 0, so let us assume that S p ´aµ 0. If 2 p ¹ a and ³´gµ 0 mod p, then g 0 mod p, and hence ³¼ ´gµ 2g 0 mod p. 27) implies the reverse inequality. If 2 p a, then there is precisely one root of ³ modulo p, by (i), but there may be none or more than one modulo pe , as in the examples a p and a 0, respectively, when e 1.