By Richard Bellman

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**Extra info for Analytic Number Theory: An Introduction**

**Example text**

Prove that the number of solutions of the congruence f (x) ≡ 0 (mod p) is 1 + dp , where d = b2 − 4ac and = 0 if p divides d. (xiii) Find the number of solutions (mod 997) of (a) x 2 + x + 1 ≡ 0, (b) x 2 + x − 2 ≡ 0, (c) x 2 + 25x − 93 ≡ 0. (xiv) With the notation of Exercise (xii), show that, if p does not divide d, then d p p x =1 f (x) p =− a . p Evaluate the sum when p divides d. (xv) Prove that if p is a prime ≡ 1 (mod 4) and if p = 2 p + 1 is a prime then 2 is a primitive root (mod p). For which primes p with p = 2 p + 1 prime is 5 a primitive root (mod p)?

We have g p−1 = 1 + py for some integer y and so, by the binomial theorem, g p−1 = 1 + pz, where z ≡ y + ( p − 1)g p−2 x (mod p). The coeﬃcient of x is not divisible by p and so we can choose x such that (z, p) = 1. Then g has the required property. For suppose that g belongs to d (mod p j ). Then d divides φ( p j ) = p j−1 ( p − 1). But g is a primitive root (mod p) and thus p − 1 divides d. Hence d = p k ( p − 1) for some k < j. Further, since p is odd, we have k (1 + pz) p = 1 + p k+1 z k , where (z k , p) = 1.

The result follows. Let g be a primitive root (mod p). We prove now that there exists an integer x such that g = g + px is a primitive root (mod p j ) for all prime powers p j . We have g p−1 = 1 + py for some integer y and so, by the binomial theorem, g p−1 = 1 + pz, where z ≡ y + ( p − 1)g p−2 x (mod p). The coeﬃcient of x is not divisible by p and so we can choose x such that (z, p) = 1. Then g has the required property. For suppose that g belongs to d (mod p j ). Then d divides φ( p j ) = p j−1 ( p − 1).