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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 . n ≥ 1 ⇒ 2n ≥ 2 ⇒ 4n ≥ 2n + 2 ⇒ n ≥ π(2n) ≥ 2n + 2 log2 log2 x ≥ 4 logx 4 logx Therefore a= log2 4 2n+2 4 . 38) log2 where n > 1. Let 2n = 2r for r ≥ 3. 1. Hence 22j+1 j π(22j ) 4 ≤ 22j j Clearly this also holds for j = 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 . n ≥ 1 ⇒ 2n ≥ 2 ⇒ 4n ≥ 2n + 2 ⇒ n ≥ π(2n) ≥ 2n + 2 log2 log2 x ≥ 4 logx 4 logx Therefore a= log2 4 2n+2 4 . 38) log2 where n > 1. Let 2n = 2r for r ≥ 3. 1.

1 which gives a constructive guideline for solving any given LDE. 2 Let a, b, n be integers. 6) if m|(a − b). Properties of linear congruences 1. a1 ≡m b1 ∧ a2 ≡m b2 ⇒ a1 ± a2 ≡m b1 ± b2 2. a1 ≡m b1 ∧ a2 ≡m b2 ⇒ a1 a2 ≡m b1 b2 3. ac ≡m bc ⇒ a ≡m b where m = m/gcd(c, m) 4. Given a fixed integer m, for each integer a, there is an integer r, such that 0 ≤ r < m and a ≡m r. These properties can be easily proved by expressing a ≡m b as a = b + km. We prove Property 4 which leads to some interesting results.

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Algorithmic number theory by Arun-Kumar S.


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