By Arun-Kumar S.

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4 f (x) = n i i=1 ai x , where an = 0 then the kth derivative of f is a polynomial with degree ≤ n − k. 5 tailers expansion of f (x + h) is f (x) + hf (x) + h2 2! f (x) + · · · + hn n n! f (x), as f t (x) = 0 when t > n. 6 solving f (x) ≡pα 0 Proof: if r is a solution to f (x) ≡pα 0 then f (r) ≡pt 0 for t = 1, 2, . . , α. ji consider α ≥ 2. if there is a solution uiα of f (x) ≡pα 0 then there is solution uα−1 of f (x) ≡pα−1 0 such that ji α−1 i for some integer v. 3) i but f (ujα−1 ) ≡pα−1 0. 4 we can find all the solutions of v and then ujα−1 +vpα−1 α will be solutions of f (x) ≡p 0 i .

Ak−1 , ak + = = and so the result holds for all n. 2 If x > 1 and x + 1/x < i. x < α = ii. 1 x Proof: > −β √ 5 then √ 5+1 2 √ = 5−1 2 √ Note that α and β are roots of equation x + 1/x = 5. √ x + 1/x < 5 ⇒ (x − α)(x − β) < 0 The two possibilities are α < x < −β) or −β < x < α. 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.

Therefore, we arrive at a contradiction and our initial assuption that there are only a finite number of primes does not hold. 5 If pi is the ith prime number, with p1 = 2, we can claim that pm+1 ≤ p since there is a prime factor of p that is not covered in p1 , p2 , . , pm . 6 If the pn denotes the nth prime, then pn ≤ 22 n−1 (the first prime p1 = 2). Proof: We present a proof by induction on n. Induction Hypothesis: For all n ≤ k, if pn denotes the nth n−1 0 n−1 prime, then pn ≤ 22 . Base Case: If n = 1, then pn = 2, and 22 = 22 = 2, hence 2 ≤ 2.

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

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