By Willem de Graaf

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12. Factorise n = 115 with the elliptic curve method. ) Chapter 3 Polynomial Factorisation The problem considered in this chapter is to find the factorisation of a polynomial in k[x], where k is a field. The algorithms depend heavily on k. Here we treat algorithms for finite fields k = Fq , with q = pn , and k = Q. 1 For f, g ∈ k[x] there exist unique q, r ∈ k[x] with deg(r) < deg(g) and f = qg + r (division with remainder). 2 Let I ⊂ k[x] be an ideal. Then I is generated by a single element, that is, there is a g ∈ k[x] with I = {f g | f ∈ k[x]}.

G5 ). (b) Prove that g4 , g5 , g6 generate I, and that they form a Gr¨ obner basis of I. (c) Compute a Gr¨ obner basis of I ∩ C[y, z] and of I ∩ C[z]. (d) Find all solutions of g1 = g2 = g3 = 0. 9. Let I ⊂ C[x, y, z] be the ideal generated by g1 = x2 yz − yz − x, g2 = xy 2 z − xy − y, g3 = xyz 2 − xy − z. Let f1 = x − z 4 + z 2 , f2 = y − z, f3 = z 7 − 2z 5 + z 3 − z. (a) Prove that I = f1 , f2 , f3 (hint: use the result of exercise 6). (b) Show that {f1 , f2 , f3 } is a Gr¨ obner basis of I with respect to the order

We have n − 1 = 26 · 27. So we take m = 27. Let x = 3. Then x27 = 664 mod n and x27·2 = 1 mod n. Hence the Miller-Rabin test shows that n is not prime. 1 Factorisation The method of Fermat Note that we may assume that n is odd. Set A = {(a, b) | 0 < a ≤ b and n = ab} and B = {(s, t) | 0 ≤ s < t and n = t2 − s2 }. b+a There is a bijection σ : A → B with σ(a, b) = b−a 2 , 2 . Indeed: b−a 2 b+a 2 − = ab = n 2 2 and σ −1 (s, t) = (t − s, t + s). The conclusion is that finding a factorisation n = ab is equivalent to finding s and t with n = t2 − s2 .

### Computational Algebra by Willem de Graaf

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