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Am1 , am2 , am3 , . . , amφ(n) is also reduced residue system modulo n. hence corrosponding to each mi there is one and only one amj such that mi ≡n amj So from previous lemma, am1 , am2 , am3 , . . , amφ(n) are congruent, not necessarily in order of appearance, to m1 , m2 , m3 , . . , mφ(n) So on taking the product of these φ(n) congruences, we get - ⇒ a φ(n) i=1 φ(n) ami φ(n) ≡n i=1 mi φ(n) m i ≡n i=1 mi aφ(n) ≡n 1 φ(n) i=1 since gcd(mi , n) = 1 and mi has inverse modulo n , so we cancel out this from both side.

2 k 22 −1 + 1 Summing up 2i 2k 2 ✷ And that completes the proof. 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 , .

Further, no qi is a factor of N . Therefore, N has a factor that is of the form 4n + 3 other than the qi for 1 ≤ i ≤ k. But by our assumption qi are the only prime numbers of the form 4n + 3. This brings us to a contradiction and hence there are an infinite number of primes of the form 4n + 3. ✷ Generalizing, we may wish to ask if there are any primes of a general form a + ib, where a and b are integers and i ranges over the naturals. 9 If the n terms of the arithmetic progression p, p + d, p + 2d, .

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Algorithmic number theory by Arun-Kumar S.

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