By Arun-Kumar S.

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P − 1)a. Claim: Any two distinct numbers from the above sequence are incongruent modulo p. Take any two numbers from the sequence, say ia and ja where i < j. Then, ia ≡p ja ⇒ p|(j − i) since p doesnt’t divide a. But 1 ≤ i < j < p, so p cannot divide j − i. Hence ia and ja are incongruent modulo p. 18) ia ≡p j where, 1 ≤ j < p and j is determined uniquely by i. Multiplying Eq. 2 . . 19) (p − 1)! ap−1 p−1 a ap ≡p ≡p ≡p Note that when we vary i in the LHS of Eq. 18, we get a different value of j each time.

Combining the last two inequalities, we get an < 1, which is a contradiction and the claim is proved. Since an irrational has infinite convergents, Hurwitz’s theorem follows from the claim. 4 For any constant c > √ 5 , Hurwitz’s theorem does not hold. ✷ 32 CHAPTER 6. RATIONAL APPROXIMATION OF IRRATIONALS Proof: Consider the irrational number α = [1, 1 . ]. There exists n ≥ 0 such that, αn = α ,pn = Fn and qn = Fn−1 . qn 1 qn ) = lim ( ) = = −β lim ( n→inf pn n→inf qn+1 α |α− pn | qn 1 = qn−1 (αn qn−1 + qn−2 ) 1 qn2 (αn+1 + qn−1 qn ) = Consider the term αn+1 + qn−1 qn .

7 There are at least n + 1 primes that are less than 22 . 1 The product of any two terms of the form 4n + 1 is also of the form 4n + 1. 40 CHAPTER 8. PRIMES AND THER INFINITUDE Proof: Consider n1 = 4k1 +1 and n2 = 4k2 +1. Therefore n1 n2 = (4k1 +1)(4k2 +1) = 16k1 k2 +4(k1 +k2 )+1 = ✷ 4k + 1 with k = 4k1 k2 + (k1 + k2 ). 8 There are an infinite number of primes of the form 4n + 3. Proof: We present a proof by contradiction. Let us assume that q1 , q2 , . , qk are the only primes that are of the form 4n + 3.

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

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