0
$\begingroup$

From HYPOTHESIS H AND AN IMPOSSIBILITY THEOREM OF RAM MURTY.

On p. 13

BUNYAKOVSKY’S CONJECTURE ( WEAK FORM ). Let $f$ be a polynomial with integer coefficients and positive leading coefficients which is irreducible over $\mathbf{Q}$. Let $d := \gcd\{ f(n)\}_{n \in \mathbf{Z}} $. Then $f (n)/d$ is prime for at least one positive integer $n$.

On p. 6:

BUNYAKOVSKY’S CONJECTURE. Let $f$ be a polynomial with integer coefficients and positive leading coefficients which is irreducible over $\mathbf{Q}$. Let $d := \gcd\{ f(n)\}_{n \in \mathbf{Z}} $. Then $f (n)/d$ is prime for infinitely many positive integers $n$.

The standard formulation of Bunyakovsky's conjecture requires $d=1$.

We claim explicit counterexample to both.

Let $f$ be the degree $12$ polynomial:

f=x^12 - 197*x^11 + 16976*x^10 - 859146*x^9 + 28693351*x^8 - 669930367*x^7 + 11254469458*x^6 - 137432686432*x^5 + 1213054497367*x^4 - 7558495778147*x^3 + 31594161981276*x^2 - 79613982402450*x + 91528191555876

$$f=x^{12} - 197 x^{11} + 16976 x^{10} - 859146 x^{9} + 28693351 x^{8} - 669930367 x^{7} + 11254469458 x^{6} - 137432686432 x^{5} + 1213054497367 x^{4} - 7558495778147 x^{3} + 31594161981276 x^{2} - 79613982402450 x + 91528191555876$$

$f$ is irreducible over $\mathbf{Q}[x]$. $\gcd(f(1),f(7))=6$, so $d \le 6$ and by congruence arguments $d=6$.

Let $N=546=2 \cdot 3 \cdot 7 \cdot 13$.

For all natural, $x$ we claim $\gcd(f(x)/6,N) > 1$. $f(x)/6$ is periodic modulo $N$, so must check only $x \in [1,N]$.

By fast computation, $\gcd(f(x)/6,N) > 1$ for $x \in [1,N]$.

So $f(x)/6$ can prime only if it equals the primes factors of $N$ at positive integers, which is not possible even at rationals.

Q1 What is wrong with this alleged counterexample?

In case of positive answer:

Q2 From it, can we can counterexample with $d=1$, which is the usual formulation of Bunyakovsky's conjecture.

Searching the web, couldn't find how to contact the author of the paper.

Q3 How to contact the author of the paper via email?

Got numeric support in both sage and pari.

Trying to follow the proof of the reformulation, we couldn't find $A,B$ with the desired properties.

Sage verification code:

def bunyakovskweakyani1():
    """
    """
    K.<x>=QQ[]
    f=x^12 - 197*x^11 + 16976*x^10 - 859146*x^9 + 28693351*x^8 - 669930367*x^7 + 11254469458*x^6 - 137432686432*x^5 + 1213054497367*x^4 - 7558495778147*x^3 + 31594161981276*x^2 - 79613982402450*x + 91528191555876
    N=546
    print 'factor(f)',factor(f)
    d=gcd(ZZ(f(1)),ZZ(f(7)))
    gg=gcd([ZZ(f(i)) for i in xrange(N)])
    print 'gcd(f(1),f(7))=d=',d,'g=',gg
    print '  should not raise error'
    for i in xrange(N):
        a=ZZ(f(i)/d)
        g=gcd(a,N)
        if g == 1:
            print 'g == 1',i
            assert False,'coprime to N'
    print '  no error, all roots should be non-integers'
    for p,_ in factor(N):
        ro=(f-p).roots(multiplicities=False)
        print p,'roots=',ro
$\endgroup$
1
  • 2
    $\begingroup$ I do not understand the comment: "searching the web, couldn't find out how to contact the author of the paper". If you delete the last part of your URL, you are led directly to the author's website, which lists his email address. $\endgroup$ Dec 9, 2015 at 19:43

2 Answers 2

11
$\begingroup$

$f(x)/6 + N\mathbb Z$ is not periodic modulo $N$. It is periodic modulo $6N$, so you have to check a larger range. And indeed, $\gcd(f(637)/6,N)=1$.

$\endgroup$
1
  • $\begingroup$ Indeed, thank you :) Is the reformulation in the paper correct? $\endgroup$
    – joro
    Dec 9, 2015 at 12:30
4
$\begingroup$

Peter Mueller has already pointed out the problem in your argumentation that your polynomial $f$ would be a counterexample to Bunyakovsky's conjecture. It is indeed not a counterexample at least to the conjecture in its weak form since $f(x)/6$ takes a prime value e.g. for $x = 6293$.

$\endgroup$
2
  • 6
    $\begingroup$ And it's also prime at 15029, 19397, 23569, 25025, 31045, 32137, and 43225. I am surprised that before the OP posted the question no reality check was made by testing this into the tens of thousands, which can be done in less than 5 seconds with a computer. $\endgroup$
    – KConrad
    Dec 9, 2015 at 15:02
  • 1
    $\begingroup$ @KConrad: Given the degree, the size of the coefficients of the polynomial and the many $f(x)/6$ which have small divisors, in fact I think one probably should choose even a vastly larger test range before only thinking it might be a counterexample. $\endgroup$
    – Stefan Kohl
    Dec 9, 2015 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.