By Jan-Hendrik Evertse

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**Example text**

6. Let a, b be integers with a p n . pj 2b > 0. Then divides a−b+1 p a a . b Proof. We have a b = a(a − 1) · · · (a − b + 1) , 1 · 2···b Hence any prime with a − b + 1 inator. p a − b + 1 > b. 7. Let a, b be integers with a > b > 0. Suppose that some prime power pk divides ab . Then pk a. Proof. Let p be a prime. 5 we have ordp a b a! (a − b)! = ordp ∞ a−b b a − − j j j p p p = j=1 . Each summand is either 0 or 1. Further, each summand with pj > a is 0. Hence ordp ab α, where α is the largest j with pj a.

Then π(x) ∼ x as x → ∞. log x This result was conjectured by Legendre in 1798. In 1851/52, Chebyshev proved that if the limit limx→∞ π(x) log x/x exists, then it must be equal to 1, but he couldn’t prove the existence of the limit. 056 . log x log x ¨ In 1859, Riemann published a very influential paper (B. Riemann, Uber die Anzahl der Primzahlen unter einer gegebenen Große, Monatshefte der Berliner Akademie der Wissenschaften 1859, 671–680; also in Gesammelte Werke, Leipzig 1892, 145–153) in which he related the distribution of prime numbers to properties 25 of the function in the complex variable s, ∞ n−s ζ(s) = n=1 (nowadays called the Riemann zeta function).

16. Let z0 ∈ C, r > 0 and let f : D0 (z0 , r) → C be analytic. Assume that z0 is either a removable singularity or a pole of f . Then z0 is a simple pole or (if z0 is neither a zero nor a pole of f ) a removable singularity of f /f , and res(z0 , f /f ) = ordz0 (f ). Proof. Let ordz0 (f ) = k. This means that f (z) = (z − z0 )k g(z) with g analytic around z0 and g(z0 ) = 0. Consequently, f (z − z0 ) g k g =k + = + . f z − z0 g z − z0 g 49 The function g /g is analytic around z0 since g(z0 ) = 0.