By Ivan Morton Niven, Herbert S. Zuckerman, Hugh L. Montgomery
The 5th version of 1 of the traditional works on quantity idea, written through internationally-recognized mathematicians. Chapters are fairly self-contained for larger flexibility. New beneficial properties contain elevated therapy of the binomial theorem, innovations of numerical calculation and a bit on public key cryptography. comprises a superb set of difficulties.
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Extra resources for An Introduction to the Theory of Numbers, 5th Edition
Of (a 2 , b 3 )? 14. Evaluate (ab, p4) and (a + b, p4) given that (a, p2) = P and (b, p3) = p2 where p is a prime. 15. 6), what conditions must be satisfied by the exponents if a is to be a cube? For a 2 1b 2 ? 16. Find an integer n such that n/2 is a square, n/3 is a cube, and n/5 is a fifth power. 17. Twin primes are those differing by 2. Show that 5 is the only prime belonging to two such pairs. Show also that there is a one-to-one correspondence between twin primes and numbers n such that n 2 - 1 has just four positive divisors.
X permutations and equating this to n! we find that X = (~). 7T, n! We now see that the quotient k! (n _ k)! is an integer, because it represents the number of ways of doing something. In this way, combinatorial interpretations can be useful in number theory. 6. 21 The product of any k consecutive integers is divisible by k!. Proof Write the product as n(n - 1) ... (n - k + 1). If n ;;;. 20. If 0 ~ n < k, then one of the factors of our product is 0, so the product vanishes, and is therefore a multiple of k!
The factoring ofany integer n > 1 into primes is unique apart from the order of the prime factors. First Proof Suppose that there is an integer n with two different factorings. 3) 24 Divisibility where the factors Pi and qj are primes, not necessarily all distinct, but where no prime on the left side occurs on the right side. But this is impossible because P l lqlq2 ... 15, PI is a divisor of at least one of the Qj. 4) It is clear that rand s are greater than 1. Now the primes PI> P2" . 4) to get two distinct factorings of n/PI' But this would contradict our assumption that all integers smaller than n are uniquely factorable.