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By Arun-Kumar S.

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N - kφ(n) ) = φ(n) n - ( k1 + k2 + . . + kφ(n) ) 1 So k∈φ(n) k = k∈φ(n) (n − k) = φ(n)n − k∈φ(n) k. 6 these n = 30 , φ(30) = 8 less than 30 and are relatively prime to 30 . 6 8 integers { 1, 7, 11, 13, 17, 19, 23, 29} are {1, 7, 11, 13, 17, 19, 23, 29} = 120 = 12 30 8 Different Proof of CRT Euler s generalisation of Fermat Little Theorem leads to a different proof of Chinese Remainder Theorem. if gcd (mi , mj ) = 1 for i = j . Then system of linear congruences x ≡mi ai ,for i = 1,2, . .

M|f (u) implies that piαi |f (u) for each i 63 −1 i α and the mi bi aiji where mi = m/pα i and bi ≡pi i mi 64 CHAPTER 12. CONGRUNCES OF HIGHER DEGREE i α 3. for each i if pα i |f (u) implies that f(u) ≡p i 0 i ✷ Proof for the second claim is very similar to the above and it can be easily proven. Now we will prove our third claim. Proof: 1. ) 2. ) 3. f (u) ≡pαi f (ai ) ≡pαi 0 from the fact that ai is a solution f (u) ≡pαi 0. i i i i 4. it means that ∀i pα i |f (u). 5. k i=1 i pα i |f (u) implies that m|f (u) 6.

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)! 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, .

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