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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Bunyakovsky conjecture ： ウィキペディア英語版 Bunyakovsky conjecture The Bunyakovsky conjecture (or Bouniakowsky conjecture) stated in 1857 by the Russian mathematician Viktor Bunyakovsky, asserts when a polynomial $f(x)$ in one variable with positive degree and integer coefficients should have infinitely many prime values for positive integer inputs. Three necessary conditions are # the leading coefficient of $f(x)$ is positive, # the polynomial is irreducible over the integers, and # as $n$ runs over the positive integers, the numbers $f(n)$ should be relatively prime. (Thus, the coefficients of $f(x)$ should be relatively prime.) Bunyakovsky's conjecture is that these three conditions are sufficient: if $f(x)$ satisfies the three conditions then $f(n)$ is prime for infinitely many positive integers $n$. For example, all cyclotomic polynomials are irreducible, with positive (in fact, =1) leading coefficient, and as $x$ runs over the positive integers, $\backslash Phi\_n(x)$ don't share a common factor greater than 1. Thus, all cyclotomic polynomials are in Bunyakovsky's conjecture, so it is conjectured strongly that for all natural number ''n'', there are infinitely many natural number ''x'' such that $\backslash Phi\_n(x)$ is prime. In fact, it can be shown that if for all natural number ''n'', there exists a natural number ''x'' > 1 such that $\backslash Phi\_n(x)$ is prime, then for all natural number ''n'', there are infinitely many natural number ''x'' such that $\backslash Phi\_n(x)$ is prime. The smallest natural number ''x'' > 1 such that $\backslash Phi\_n(x)$ is prime are :3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 6, 2, 4, 3, 2, 10, 2, 22, 2, 2, 4, 6, 2, 2, 2, 2, 2, 14, 3, 61, 2, 10, 2, 14, 2, 15, 25, 11, 2, 5, 5, 2, 6, 30, 11, 24, 7, 7, 2, 5, 7, 19, 3, 2, 2, 3, 30, 2, 9, 46, 85, 2, 3, 3, 3, 11, 16, 59, 7, 2, 2, 22, 2, 21, 61, 41, 7, 2, 2, 8, 5, 2, 2, ... (It is conjectured strongly that all the terms of this sequence are defined. However, some terms are very large, for example, the 545th term of this sequence is 2706, the 601st term of this sequence is 2061, and the 943rd term of this sequence is 2042.) We need the first condition because if the leading coefficient is negative then for all large $x$, and thus $f(n)$ is not a prime number for large positive integers $n$. If we allow negative primes like -2, -3, -5, ... to count as prime numbers then this first condition can be dropped; the second and third necessary conditions are more substantial, as we will see below, since they imply $f(n)$ can be prime only finitely many times for reasons that are more serious than a sign problem. We need the second condition because if $f(x)\; =\; g(x)h(x)$ where the polynomials $g(x)$ and $h(x)$ have integral coefficients and are not $\backslash pm\; 1$ then we have $f(n)\; =\; g(n)h(n)$ for all integers $n$, so $f(n)$ is composite for all large $n$ (because $g(x)$ and $h(x)$ take the values 0 and $\backslash pm\; 1$ only finitely many times). The third condition, that the numbers $f(n)$ have gcd 1, is the most technical sounding and is best understood by an example where it does not hold. Consider the polynomial $f(x)\; =\; x^2\; +\; x\; +\; 2$. It has a positive leading coefficient and is irreducible, but $f(n)$ is ''even'' for all integers $n$, so the values of this polynomial are prime only finitely many times on the positive integers (namely when it takes the value 2, which is actually only at $n\; =\; 1$ among positive integers). In practice, the easiest way to verify the third condition for a polynomial $f(x)$ is to find one pair of positive integers $m$ and $n$ such that $f(m)$ and $f(n)$ are relatively prime: when this happens no integer greater than 1 can divide all values of $f(x)$ on the positive integers because it would have to divide $f(m)$ and $f(n)$. An example of Bunyakovsky's conjecture is the polynomial ''f''(''x'') = ''x''2 + 1, for which some of the prime values that it has on positive integers are listed below. (sequence (''x'') and (''x''2 + 1) in OEIS) That $n^2+1$ should be prime infinitely often is a problem first raised by Euler, and it is also the fifth Hardy–Littlewood conjecture and the fourth of Landau's problems. The third condition in Bunyakovsky's conjecture is saying that the set of integers $f(1),f(2),\; f(3),\backslash dots$ has gcd 1. It is a surprise to most people at first that this is ''not'' the same as saying the coefficients of $f(x)$ are relatively prime together, but the example of $x^2\; -\; x\; +\; 2$ shows this. If the third condition in Bunyakovsky's conjecture holds then necessarily the coefficients of the polynomial are relatively prime (in fact, if the second condition holds then also the coefficients are relatively prime, since a common factor of the coefficients that is greater than 1 would mean the polynomial is reducible over the integers), but the converse is not true. As noted above, a practical way to prove the numbers $f(1),f(2),\; f(3),\backslash dots$ have gcd 1 is to find a single pair of values that are relatively prime. A way of calculating the gcd of all the numbers $f(n)$ when $n\; \backslash geq\; 1$, even in the case of this number being greater than 1, is to rewrite $f(x)\; =\; c\_0\; +\; c\_1x\; +\; \backslash cdots\; +\; c\_dx^d$ as a linear combination of the binomial coefficient polynomials $\backslash binom$: $f(x)\; =\; a\_0\; +\; a\_1\backslash binom\; +\; \backslash cdots\; +\; a\_d\backslash binom$. If each $c\_i$ is an integer then each $a\_i$ is an integer and $\backslash gcd\backslash \; =\; \backslash gcd(a\_0,a\_1,\backslash dots,a\_d).$ For example, $x^2\; -\; x\; +\; 2\; =\; 2\backslash binom\; +\; 2$, and the coefficients in the second formula have gcd 2, which is related to the fact that $x^2\; -\; x\; +\; 2$ has even values on the integers. Using this gcd formula it can be proved $\backslash gcd\backslash$ is 1 if and only if there is some pair of positive integers $m$ and $n$ such that $f(m)$ and $f(n)$ are relatively prime. To date, the only case of Bunyakovsky's conjecture that has been proved is polynomials of degree 1. This is Dirichlet's theorem, which states that when $a$ and $m$ are relatively prime integers there are infinitely many prime numbers $p\; \backslash equiv\; a\; \backslash !\backslash !\; \backslash pmod\; m$. This is Bunyakovsky's conjecture for $f(x)\; =\; a\; +\; mx$ (or $a\; -\; mx$ if ). The third necessary condition in Bunyakovsky's conjecture for a linear polynomial $mx\; +\; a$ is equivalent to $a$ and $m$ being relatively prime. No single case of Bunyakovsky's conjecture for degree greater than 1 is proved, although numerical evidence in higher degree is consistent with the conjecture. ==Generalized Bunyakovsky conjecture== Given ''n'' polynomials, (''n'' can be any natural number, when ''n'' = 1, this is the original conjecture of Bunyakovsky) that each satisfy all three conditions, and for any prime ''p'' there is an ''x'' such that the values of all ''n'' polynomials at ''x'' are not divisible by ''p'' (thus, the set of polynomials: does not work since one of the values of the polynomials must be divisible by 3 for any ''x'', and neither does the set since one of the values of the polynomials must be divisible by 3 for any ''x''.), then there are infinitely many positive integers ''x'' such that all values of these ''n'' polynomials at ''x'' are prime. For example, if this conjecture is true then there are infinitely many positive integers ''x'' such that ''x''2 + 1, 3''x'' - 1, and ''x''2 + ''x'' + 41 are all prime. This conjecture includes as special cases the twin prime conjecture (when ''n'' = 2, and the two polynomials are ''x'' and ''x'' + 2) as well as the infinitude of prime quadruplets (when ''n'' = 4, and the four polynomials are ''x'', ''x'' + 2, ''x'' + 6, and ''x'' + 8), sexy primes (when ''n'' = 2, and the two polynomials are ''x'' and ''x'' + 6), Sophie Germain primes (when ''n'' = 2, and the two polynomials are ''x'' and 2''x'' + 1), and Polignac's conjecture (when ''n'' = 2, and the two polynomials are ''x'' and ''x'' + ''k'', with ''k'' any even number). When all the polynomials have degree 1 this is Dickson's conjecture. In fact, this conjecture is the same as the Generalized Dickson conjecture. With an exception of Dirichlet's theorem (the set contains only one polynomial, and it has degree 1), even there is no other single case of set of polynomials in this conjecture having been proved, including , , and . 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Bunyakovsky conjecture」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.