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By Arun-Kumar S.

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E. for every positive integer k, there exist k consecutive composite members. Proof: This can be easily seen as ∀ positive integers k we have (k + 1)! + 2, . . , (k + 1)! + k + 1. 1) j | (k + 1)! + j, ∀j ∈ 2, . . 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)!

The first one is ruled out as we are given that x > 1 > −β. So, we√have −β < x < α which proves the first claim. √ 5−1 1 √2 = ✷ Now, x < α ⇒ x < 5+1 ⇒ > which proves the second claim. 1) Proof: We first prove certain claims √ Claim. 1 is false for any consecutive Cn−1 and Cn , then rn + 1/rn < 5 where rn = qn /qn−1 . n−1 1 . So, | x − Cn−1 | + | x − Cn | ≥ √15 ( q12 + We are given | x − pqn−1 | ≥ √5q12 and | x − pqnn | ≥ √5q 2 n n−1 x lies between Cn−1 and Cn , | x − Cn−1 | + | x − Cn | = | ⇒ 1 qn−1 qn qn qn−1 ≥ ≥ ⇒ rn ⇒ rn + 1/rn ≥ ≤ pn qn − pn−1 qn−1 n |= 1 qn−1 qn .

Prkr − pkr r −1 ) φ (n) = n (1 − 1 p1 ) (1 − 1 p2 ) . . (1 − 1 pr ) Proof By Induction on r , the number of distinct prime factors of n . It is true for r = 1, ki+1 since gcd ( pk11 pk22 . . pki i , pi+1 ) = 1. φ(p1k1 ) = (pk11 − pk11 −1 ) . Let it holds for r = i, by definition of multiplicative function k k i+1 i+1 ) ) = φ(pk11 . . pki i ) φ(pi+1 φ((pk11 pk22 . . pki i )pi+1 k k i+1 i+1 = φ(pk11 . . piki ) (pi+1 - pi+1 −1 ) Invoking the induction assumption first factor on right hand side becomes k k k i+1 i+1 i+1 φ(pk11 .

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Algorithmic number theory by Arun-Kumar S.


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