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

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

The 5th version of 1 of the traditional works on quantity concept, written via internationally-recognized mathematicians. Chapters are fairly self-contained for better flexibility. New good points comprise extended remedy of the binomial theorem, recommendations of numerical calculation and a bit on public key cryptography. comprises an exceptional set of difficulties.

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Extra resources for An Introduction to the Theory of Numbers, 5th Edition

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8 here. 1 If an integer m, not zero, divides the difference a - b, we say that a is congruent to b modulo m and write a == b (mod m). If a - b is not divisible by m, we say that a is not congruent to b modulo m, and in this case we write a ¥ b (mod m). Since a - b is divisible by m if and only if a - b is divisible by -m, we can generally confine our attention to a positive modulus. Indeed, we shall assume throughout the present chapter that the modulus m is a positive integer. Congruences have many properties in common with equalities.

3, part 2, shows that arj == arj (mod m) implies rj == rj (mod m) and hence i = j. 7 Fermat's theorem. Let p denote a prime. If p,(a then a P - 1 == Hmod p). For every integer a, a P == a (mod p). 8. 8 then Euler's generalization of Fermat's theorem. If (a, m) = 1, acf>(m) == 1 (mod m). Proof Let r l , r z ,"', rcf>(m) be a reduced residue system modulo m. 6, ar l , arz ," " arcf>(m) is also a reduced residue system modulo m. Hence, corresponding to each rj there is one and only one arj such that rj == arj (mod m).

13) for Izl < 1. Here a is an arbitrary real or complex number. 12) if a is a non-negative integer. As a function of a, the quantity (~) is a polynomial of degree k with rational coefficients. 21 we see that this polynomial takes integral values whenever a is an integer. A polynomial with this property is called integer-valued. 13) is the Taylor series of the function on the left. To demonstrate that it converges to the desired value, one may use the integral form of the remainder, which states that if f(z) is a function for which f(K+l)(Z) is continuous, then where We take f(z) = 0 + z)"', so that f(k)(Z) = a(a -1) ...

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