By Arun-Kumar S.

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Let (x0 , y0 ) be a particular solution and (x , y ) be any other solution of the LDE. 5) 52CHAPTER 10. t x = x0 − su = x0 − (b/d)u and y = y0 + ru = y0 + (a/d)u. ✷ We now give a procedure that computes a particular solution for the given LDE. All the other solutions can be derived using this particular solution. 1 Solving a Linear Diophantine Equation P rocedure(LDE(ax + by = c)) Let (d, x , y ) = ExtendedEuclid(a, b). If d|c then x0 ← cx /d y0 ← cy /d return (x0 , y0 ) else print ”No solutions” EndP roc.

1 pα || n means pα | n but pα+1 | n 45 46 CHAPTER 9. 7 If for prime p and n ≥ 1 pα || n! Clearly n = 0 and n = 1 are trivial cases. Therefore we ∞ β= i=1 n−1 and pβ || (n − 1)! 7) k ✷ We therefore have α = β + k where p || n and hence since n! = n(n − 1)! and from above we have pβ || (n − 1)! therefore pα || n! 8 For all m, n prime p for pα || n! m! 10) 2n! n! 1. ; p | n! 28) ✷ 48 CHAPTER 9. 2 We have 2n n a= log2 4 ≤ 2n But since 2n n and since for j ∈ {1, 2, . . 34) for x ≥ 2, choose n such that 2n ≤ x < 2n + 2 .

T. sn+1 < d √ ⇒ 0 ≤ |mn+1 | < d f orall j < k mi = mj so that sj = sk and x = [a0 , . . , aj−1 , aj , . . 2 Every SPCF has quadratic representation. P roof : First suppose that [a0 , a1 , . . , an , an+1 , . . , an+k ] is a periodic continued fraction. Set α = [an+1 , an+2 , . ]. Then α = [an+1 , . . , an+k , α], so α= αpn+k +pn+k−1 αqn+k +qn+k−1 . ) Thus α satisfies a quadratic equation. Since the ai are all integers, the number [a0 , a1 , . ] = = [a0 , a1 , . . , an , α] 1 a0 + 1 a1 + a2 +···+α can be expressed as a polynomial in α with rational coefficients, so [a0 , a1 , .

### Algorithmic number theory by Arun-Kumar S.

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