By Pierre Samuel

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See Dickson’s conjecture; Dirichlet; Hardy-Littlewood conjectures Aurifeuillian factorization Since a2 + b2 cannot be factorized into two algebraic factors, unlike a2 − b2 = (a + b)(a − b), we might assume that n4 + 1, which is also the sum of two squares, cannot be factorized. Not so! qxd 3/22/05 12:05 PM Page 15 average prime • 15 ෆ+ Now we can see a connection: a2 + b2 = (a + b)2 − 4ab = (a − ͙ab ෆ + b), which normally “doesn’t count” because of the b)(a + ͙ab square roots. It follows that n4 + 1 is always composite, except when n2 − n + 1 = 1 and n = 0 or 1.

Dirichlet proved in 1837, a conjecture made by Gauss: if a and b are coprime positive integers, then the arithmetic progression a, a + b, a + 2b, a + 3b, . . contains infinitely many primes. He did so by proving that if p is a prime of the form an + b, with a and b coprime, then the sum of all the primes p of this form less than x is approximately, 1 ᎏ и log log x φ(a) as x tends to infinity. In other words, it increases without limit, albeit very slowly, and so the primes of that form cannot be finite in number.

There are no primes among the first 13,500 terms. (Weisstein, MathWorld) consecutive numbers Sylvester proved in 1892 that every product of n consecutive integers greater than n is divisible by a prime greater than n. In fact, although the product of five consecutive integers 6 и 7 и 8 и 9 и 10 is divisible by just one prime, 7, greater than 5, the product 200 и 201 и 202 и 203 и 204 is divisible by five primes, 67, 101, 7, 29, and 17. This suggests that Sylvester’s result is rather generous: indeed, the product of just two consecutive numbers is always divisible by a prime greater than N, if the product is large enough.

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Algebraic theory of numbers by Pierre Samuel


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