By Ivan Morton Niven, Herbert S. Zuckerman, Hugh L. Montgomery

The 5th variation of 1 of the normal works on quantity conception, written through internationally-recognized mathematicians. Chapters are really self-contained for larger flexibility. New positive factors comprise improved remedy of the binomial theorem, thoughts of numerical calculation and a bit on public key cryptography. includes an exceptional set of difficulties.

**Read Online or Download An Introduction to the Theory of Numbers, 5th Edition PDF**

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**Sample text**

16). Show that if the c k are integers, then P(x) is an integer-valued polynomial. 16). Show that if P(O), P(1),' . " P(n) are integers then the c k are integers and P(x) is integer-valued. Show that if f(x) is a polynomial of degree n with real coefficients, which takes integral values on a certain set of n + 1 consecutive integers, then f(x) is integer-valued. f(x) is a polynomial with integral coefficients. d. (f(O), f(1),' . " f(n». Show that glf(k) for all integers k. Show that if m and n are non-negative integers then Q(x) = t (_l)k(m k+ 1 )(m +mn- k) = k=O *20.

N + k - 1) = ( -1) k(-n+k-1) k k!. 20 the binomial coefficient is an integer. /: i ft d} containing n - k elements. The binomial theorem. For any integer n ;;;. 11) Proof We consider first the product n n (x; + y;). 4 37 The Binomial Theorem On multiplying this out, we obtain 2 n monomial terms of the form nXinYi iEN i<1=N where N is any subset of {l, 2,' . " n}. For each fixed k, 0 ~ k ~ n, we consider the monomial terms obtained from those subsets N of {l, 2,' .. , n} having exactly k elements.

P - 1, p with the exception of p are relatively prime to p. Thus we have 4J(p) = p - 1, and the first part of Fermat's theorem follows. The second part is now obvious. 9 If (a, m) = 1 then there is an x such that ax == 1 (mod m). Any two such x are congruent (mod m). If (a, m) > 1 then there is no such x. Proof If (a, m) = 1, then there exist x and y such thiit ax + my = 1. That is, ax == 1 (mod m). Conversely, if ax == 1 (mod m), then there is a y such that ax + my = 1, so that (a, m) = 1. 3 that XI == x 2 (mod m).