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Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b1110401.png" /> be a [[Polynomial|polynomial]] of degree <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b1110402.png" /> with integer coefficients. Already in 1854, V. Bunyakovskii [[#References|[a1]]] considered the problem whether <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b1110403.png" /> represents infinitely many prime numbers as <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b1110404.png" /> ranges over the positive integers (cf. [[Prime number|Prime number]]). There are some obvious necessary conditions, e.g., that the coefficients of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b1110405.png" /> are relatively prime, that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b1110406.png" /> is irreducible (cf. [[Irreducible polynomial|Irreducible polynomial]]) and, trivially, that the leading coefficient is positive. Are these conditions sufficient?
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As Bunyakovskii remarked, the answer is  "no" . For instance, for each prime number <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b1110407.png" /> one has
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<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b1110408.png" /></td> </tr></table>
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Let  $  f ( x ) $
 +
be a [[Polynomial|polynomial]] of degree  $  \geq 1 $
 +
with integer coefficients. Already in 1854, V. Bunyakovskii [[#References|[a1]]] considered the problem whether  $  f ( n ) $
 +
represents infinitely many prime numbers as  $  n $
 +
ranges over the positive integers (cf. [[Prime number|Prime number]]). There are some obvious necessary conditions, e.g., that the coefficients of  $  f $
 +
are relatively prime, that  $  f $
 +
is irreducible (cf. [[Irreducible polynomial|Irreducible polynomial]]) and, trivially, that the leading coefficient is positive. Are these conditions sufficient?
  
Replacing the constant term <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b1110409.png" /> by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104010.png" /> with a suitable integer <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104011.png" />, one can make <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104012.png" /> irreducible, say with <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104013.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104014.png" />, etc. Hence, one has to assume that the values <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104015.png" /> for positive integers <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104016.png" /> are not all divisible by a prime number. Bunyakovskii's conjecture is that these conditions are sufficient.
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As Bunyakovskii remarked, the answer is  "no" . For instance, for each prime number $  p $
 +
one has
  
A special case of this conjecture is that the polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104017.png" /> represents infinitely many prime numbers. Similarly, the [[Dirichlet theorem|Dirichlet theorem]] about infinitely many primes in an arithmetic progression comes from considering the polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104018.png" /> with relatively prime integers <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104019.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/b/b111/b111040/b11104020.png" />.
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$$
 +
n  ^ {p} - n - p \equiv0 ( { \mathop{\rm mod} } p )  \textrm{ for  all  integers  }  n.
 +
$$
 +
 
 +
Replacing the constant term  $  p $
 +
by  $  pk $
 +
with a suitable integer  $  k $,
 +
one can make  $  x  ^ {p} - x - pk $
 +
irreducible, say with  $  p = 2 $,
 +
$  p = 3 $,
 +
etc. Hence, one has to assume that the values  $  f ( n ) $
 +
for positive integers  $  n $
 +
are not all divisible by a prime number. Bunyakovskii's conjecture is that these conditions are sufficient.
 +
 
 +
A special case of this conjecture is that the polynomial $  x  ^ {2} + 1 $
 +
represents infinitely many prime numbers. Similarly, the [[Dirichlet theorem|Dirichlet theorem]] about infinitely many primes in an arithmetic progression comes from considering the polynomial $  ax + b $
 +
with relatively prime integers $  a > 0 $
 +
and b $.
  
 
Bunyakovskii's conjecture was rediscovered and generalized to several polynomials by A. Schinzel [[#References|[a2]]]; see also the comments in [[#References|[a3]]].
 
Bunyakovskii's conjecture was rediscovered and generalized to several polynomials by A. Schinzel [[#References|[a2]]]; see also the comments in [[#References|[a3]]].

Latest revision as of 06:29, 30 May 2020


Let $ f ( x ) $ be a polynomial of degree $ \geq 1 $ with integer coefficients. Already in 1854, V. Bunyakovskii [a1] considered the problem whether $ f ( n ) $ represents infinitely many prime numbers as $ n $ ranges over the positive integers (cf. Prime number). There are some obvious necessary conditions, e.g., that the coefficients of $ f $ are relatively prime, that $ f $ is irreducible (cf. Irreducible polynomial) and, trivially, that the leading coefficient is positive. Are these conditions sufficient?

As Bunyakovskii remarked, the answer is "no" . For instance, for each prime number $ p $ one has

$$ n ^ {p} - n - p \equiv0 ( { \mathop{\rm mod} } p ) \textrm{ for all integers } n. $$

Replacing the constant term $ p $ by $ pk $ with a suitable integer $ k $, one can make $ x ^ {p} - x - pk $ irreducible, say with $ p = 2 $, $ p = 3 $, etc. Hence, one has to assume that the values $ f ( n ) $ for positive integers $ n $ are not all divisible by a prime number. Bunyakovskii's conjecture is that these conditions are sufficient.

A special case of this conjecture is that the polynomial $ x ^ {2} + 1 $ represents infinitely many prime numbers. Similarly, the Dirichlet theorem about infinitely many primes in an arithmetic progression comes from considering the polynomial $ ax + b $ with relatively prime integers $ a > 0 $ and $ b $.

Bunyakovskii's conjecture was rediscovered and generalized to several polynomials by A. Schinzel [a2]; see also the comments in [a3].

P.T. Bateman and R. Horn have conjectured an asymptotic behaviour (cf. Bateman–Horn conjecture).

References

[a1] V. Bouniakowsky [V. Bunyakovskii], "Sur les diviseurs numériques invariables des fonctions rationelles entières" Mém. Sci. Math. et Phys. , VI (1854–1855) pp. 307–329
[a2] A. Schinzel, W. Sierpiński, "Sur certaines hypothèses concernant les nombres premiers" Acta Arithm. , 4 (1958) pp. 185–208
[a3] H. Halberstam, H.-E. Richert, "Sieve methods" , Acad. Press (1974)
How to Cite This Entry:
Bunyakovskii conjecture. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Bunyakovskii_conjecture&oldid=17472
This article was adapted from an original article by S. Lang (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article