Bunyakovsky conjecture explained
in one variable with
integer coefficients to give
infinitely many prime values in the sequence
It was stated in 1857 by the
Russian mathematician Viktor Bunyakovsky. The following three conditions are necessary for
to have the desired prime-producing property:
- The leading coefficient is positive,
- The polynomial is irreducible over the rationals (and integers), and
- There is no common factor for all the infinitely many values
. (In particular, the coefficients of
should be relatively prime. It is not necessary for the values f(n) to be pairwise relatively prime.) Bunyakovsky's conjecture is that these conditions are sufficient: if
satisfies (1)–(3), then
is prime for infinitely many positive integers
.
A seemingly weaker yet equivalent statement to Bunyakovsky's conjecture is that for every integer polynomial
that satisfies (1)–(3),
is prime for
at least one positive integer
: but then, since the translated polynomial
still satisfies (1)–(3), in view of the weaker statement
is prime for at least one positive integer
, so that
is indeed prime for infinitely many positive integers
. Bunyakovsky's conjecture is a special case of
Schinzel's hypothesis H, one of the most famous open problems in number theory.
Discussion of three conditions
The first condition is necessary because if the leading coefficient is negative then
for all large
, and thus
is not a (positive) prime number for large positive integers
. (This merely satisfies the sign convention that primes are positive.)
The second condition is necessary because if
where the polynomials
and
have integer coefficients, then we have
for all integers
; but
and
take the values 0 and
only finitely many times, so
is composite for all large
.
The second condition also fails for the polynomials reducible over the rationals.
P(x)=(1/12) ⋅ x4+(11/12) ⋅ x2+2
doesn't satisfy the condition (2) since
P(x)=(1/12) ⋅ (x4+11x2+24)=(1/12) ⋅ (x2+3) ⋅ (x2+8)
, so at least one of the latter two factors must be a divisor of
in order to have
prime, which holds only if
. The corresponding values are
, so these are the only such primes for integral
since all of these numbers are prime. This isn't a counterexample to Bunyakovsky conjecture since the condition (2) fails.
The third condition, that the numbers
have gcd 1, is obviously necessary, but is somewhat subtle, and is best understood by a counterexample. Consider
, which has positive leading coefficient and is irreducible, and the coefficients are relatively prime; however
is
even for all integers
, and so is prime only finitely many times (namely at
, when
).
In practice, the easiest way to verify the third condition is to find one pair of positive integers
and
such that
and
are
relatively prime. In general, for any
integer-valued polynomial
we can use
\gcd\{f(n)\}n\geq=\gcd(f(m),f(m+1),...,f(m+d))
for any integer
, so the gcd is given by values of
at any consecutive
integers.
[1] In the example above, we have
and so the gcd is
, which implies that
has even values on the integers.
Alternatively, when an integer polynomial
is written in the basis of
binomial coefficient polynomials:
each coefficient
is an integer and
\gcd\{f(n)\}n=\gcd(a0,a1,...,ad).
In the example above, this is:
and the coefficients in the right side of the equation have gcd 2.
Using this gcd formula, it can be proved
if and only if there are positive integers
and
such that
and
are relatively prime.
Examples
A simple quadratic polynomial
Some prime values of the polynomial
are listed in the following table. (Values of
form
OEIS sequence ; those of
form .)
| 1 | 2 | 4 | 6 | 10 | 14 | 16 | 20 | 24 | 26 | 36 | 40 | 54 | 56 | 66 | 74 | 84 | 90 | 94 | 110 | 116 | 120 |
---|
| 2 | 5 | 17 | 37 | 101 | 197 | 257 | 401 | 577 | 677 | 1297 | 1601 | 2917 | 3137 | 4357 | 5477 | 7057 | 8101 | 8837 | 12101 | 13457 | 14401 | |
---|
That
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. Despite the extensive numerical evidence it is not known that this sequence extends indefinitely.
Cyclotomic polynomials
The cyclotomic polynomials
for
satisfy the three conditions of Bunyakovsky's conjecture, so for all
k, there should be infinitely many natural numbers
n such that
is prime. It can be shown that if for all
k, there exists an integer
n > 1 with
prime, then for all
k, there are infinitely many natural numbers
n with
prime.
The following sequence gives the smallest natural number n > 1 such that
is prime, for
:
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, ... .
This sequence is known to contain some large terms: the 545th term is 2706, the 601st is 2061, and the 943rd is 2042. This case of Bunyakovsky's conjecture is widely believed, but again it is not known that the sequence extends indefinitely.
Usually, there is an integer
between 2 and
(where
is
Euler's totient function, so
is the
degree of
) such that
is prime, but there are exceptions; the first few are:
1, 2, 25, 37, 44, 68, 75, 82, 99, 115, 119, 125, 128, 159, 162, 179, 183, 188, 203, 213, 216, 229, 233, 243, 277, 289, 292, ....
Partial results: only Dirichlet's theorem
To date, the only case of Bunyakovsky's conjecture that has been proved is that of polynomials of degree 1. This is Dirichlet's theorem, which states that when
and
are relatively prime integers there are infinitely many prime numbers
. This is Bunyakovsky's conjecture for
(or
if
).The third condition in Bunyakovsky's conjecture for a linear polynomial
is equivalent to
and
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
See main article: Schinzel's hypothesis H.
Given
polynomials with positive degrees and integer coefficients, each satisfying the three conditions, assume that for any prime
there is an
such that none of the values of the
polynomials at
are divisible by
. Given these assumptions, it is conjectured that there are infinitely many positive integers
such that all values of these
polynomials at
are prime. This conjecture is equivalent to the generalized Dickson conjecture and
Schinzel's hypothesis H.
See also
References
Bibliography
- Rupert. Wolfgang M.. Reducibility of polynomials f(x, y) modulo p. 1998-08-05. math/9808021.
- Bouniakowsky. V.. Sur les diviseurs numériques invariables des fonctions rationnelles entières. Mém. Acad. Sc. St. Pétersbourg. 6. 305–329. 1857.
Notes and References
- Hensel . Kurt . Ueber den grössten gemeinsamen Theiler aller Zahlen, welche durch eine ganze Function von n Veränderlichen darstellbar sind . . 1896 . 1896 . 116 . 350–356 . 10.1515/crll.1896.116.350 . 118266353 .