Download An Introduction to the Theory of Numbers, 5th Edition by Ivan Morton Niven, Herbert S. Zuckerman, Hugh L. Montgomery PDF

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.

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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.

Show description

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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).

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