Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Melvyn Nathanson, in his book Elementary Methods in Number Theory (Chapter 8: Prime Numbers) states the following:

  • A conjecture of Schinzel and Sierpinski asserts that every positive rational number $x$ can be represented as a quotient of shifted primes, that $x=\frac{p+1}{q+1}$ for primes $p$ and $q$. It is known that the set of shifted primes, generates a subgroup of the multiplicative group of rational numbers of index at most $3$.

I would like to know what progress has been made regarding this problem and why is this conjecture important. Since it generates a subgroup, does the subgroup which it generates have any special properties?

I had actually posed a problem which asks us to prove that given any interval $(a,b)$ there is a rational of the form $\frac{p}{q}$ ($p,q$ primes) which lies inside $(a,b)$. Does, this problem have any connections with the actual conjecture?

I had actually posed this question on MATH.SE (Link : http://math.stackexchange.com/questions/18352/a-conjecture-of-schinzel-and-sierpinski ). I did get a decent answer from Andreis Caicedo, but i would like to have more opinions of Mathematicians from this community.

share|improve this question
2  
The question of approximating real numbers by ratios of prime numbers is interesting. One can show that for any positive real number $u$, there are infinitely many pairs of primes $(p,q)$ so that $|u - p/q|<Cq^{-0.465}$, where $C$ depends only on $u$. To prove this, you can use Baker and Harman's result; they show that there exists a prime number between $n$ and $n+n^{0.535}$. –  Hej Jan 29 '11 at 20:33
3  
That would surely do it. From the Prime Number theorem it is known that for every $\epsilon>0$ there is an $N$ so that there is always a prime in $(n,(1+\epsilon)n)$ for $n>N$. That is also enough. –  Aaron Meyerowitz Jan 29 '11 at 20:44
    
@Aaron, you are right. Actually, I am interested in knowing 'how well' we can approximate real numbers by ratios of primes, and so it seems that the result I mentioned is the best we can have without using RH. It is conjectured that for any $u>0$ and any $\epsilon>0$, there exist infinity many pairs of primes $(p,q)$ so that $|u-p/q|<\epsilon \log(q)/q$. This is definitely true for $u=1$ since "small gaps between primes exist" (see the paper by Goldston, et. al). –  Hej Jan 30 '11 at 1:37

1 Answer 1

The question can be written as follows: Given two positive integers $a$ and $b$, do there exist primes $p$ and $q$ such that $$aq-bp=b-a?$$ You would expect there to be not just one such pair of primes, but infinitely many pairs. For instance, if $a=2$ and $b=1$, then $q$ is a Sophie Germain prime, and everyone expects there to be infinitely many of those. Moreover, you should be able to replace the right side of the equation with a constant $c$, i.e., $$aq-bp=c.$$ The twin prime conjecture says that there are infinitely many solutions when $a=b=1$ and $c=2$. Polignac's conjecture implies infinitely many solutions when $a=b=1$ and for every even value of $c$. In general you should expect infinitely many solutions when there isn't some obvious congruence that forces finiteness; for instance obviously $a=b=c=1$ only has one solution. Moreover, it's natural to expect a specific slowly decreasing density of solutions using a heuristic estimate derived from the prime number theorem.

This question for all suitable $a$, $b$, and $c$ is in turn a special case of yet more general questions about linear patterns in the prime numbers. For instance, the statement that there are infinitely many arithmetic progressions of length 3 in the primes is the statement that there are infinitely many solutions to $$p-q = q - r > 0.$$ Now, it's a famous theorem of Tao and Green that there are infinitely many arithmetic progressions of primes of arbitrary length. Later Tao and Green did a more systematic study that established the existence of all kinds of linear patterns in the prime numbers. However, the Sieprinski-Schnizel conjecture, and its generalization in the previous paragraph, are part of the "rank 1 case" that they did not solve. (These are just my mental notes from a survey talk by Terry Tao that I once attended.) If they could have done the rank 1 case, it would have included the twin prime conjecture and I think that it would have implied the asymptotic Goldbach conjecture too, so that would have been even more amazing than what they did accomplish.

I have no idea whether this remaining rank 1 case is the same class of question as the Tao-Green results, but just harder; or whether it is so much harder that it is in a different class. Let's optimistically say that it's the former. If so, then what makes the Schinzel-Sierpinski conjecture interesting is that you should always expect infinitely many solutions in prime numbers to linear equations, unless there are only finitely many solutions because of a simple congruence. And I might say that the Tao-Green results are the main recent progress, even though they answered different questions.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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