Melvyn Nathanson, in his book Methods in Number Theory (Chapter 8: Prime Number's) 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 actually posted this question on MO. Interested users can see this link:

  • 4
    $\begingroup$ Any collection of nonzero rational numbers generates "a subgroup of the multiplicative group of rational numbers." In particular, "this generates a subgroup of the multiplicative group of rational numbers" requires no proof. $\endgroup$ Jan 20, 2011 at 21:31
  • 1
    $\begingroup$ @Arturo: Oh, yes. $\endgroup$
    – anonymous
    Jan 20, 2011 at 21:32

2 Answers 2


The main progress I know of is in the two papers by Elliot:

  • Elliott, P. D. T. A. "The multiplicative group of rationals generated by the shifted primes. I." J. Reine Angew. Math. 463 (1995), 169–216.
  • Elliott, P. D. T. A. "The multiplicative group of rationals generated by the shifted primes. II." J. Reine Angew. Math. 519 (2000), 59–71.

The result you mention is in the first one.

From the review at MathReviews for the first paper:

Let ${\mathbf Q}^*$ be the multiplicative group of positive rationals, let $\Gamma$ be the subgroup generated by shifted primes $p+1$ and set $G:={\mathbf Q}^*/\Gamma$. Denote by $\Gamma_k$ the subset of $\Gamma$ comprising all rationals $r$ that have a representation $r=\prod^k_{j=1}(p_j+1)^{\epsilon_j}$ with $\epsilon_j=0,\pm1$ and all $p_j$ prime with at most $k$ factors. A well-known conjecture of Schinzel and Sierpiński asserts that $\Gamma=\Gamma_2={\mathbf Q}^*$, and a weaker assumption is that $G$ is trivial. The main goal of this paper is to prove that (i) $G$ is finite; (ii) there exists an absolute $k$ such that $r^{|G|}\in\Gamma_k$ for all $r\in{\mathbf Q}^*$; (iii) $|G|\leq 3$.

The author derives (i) and (ii) by a straightforward argument from the following result, stated as Lemma 1: For $m\in{\mathbf Z}^+$, let $F_m$ be the set of positive integers representable in the form$(p+1)/(m(q+1))$ with $p,q$ prime. Then there is an absolute constant $c$ such that the lower asymptotic density $\underline{\bf d}F_m$ of $F_m$ satisfies $\underline{\bf d}F_m\geq c$ uniformly in $m$. Moreover, the same argument yields that $|G|\leq 1/c$.

If you have access to MathReviews, I recommend that you read the whole review (by Tenenbaum!), as it is very insightful. In particular, Tenenbaum mentions prior work by Meyer and himself that had given $|G|\le4$. To obtain $|G|\le 3$,

The author develops a novel way of studying $G$ based on principles of harmonic analysis. This rests on properties of multiplicative functions on the set of shifted primes. [...] The proofs rely on difficult, delicate estimates of classical analytic number theory type. The large sieve, Selberg's sieve, the dispersion method and Halász's theorem are the main ingredients, all deeply revisited.

Here is Tenenbaum's review of the second paper:

In Part I, the author proved that $G$ has order $|G|\le 3$ and that there is a $k$ such that every positive rational $r$ has a representation $r^{|G|}=\prod_{j=1}^k(p_j+1)^{\epsilon_j}$ with $p_j$ prime, $\epsilon_j=\pm1$. The standard conjecture in this context is that $G$ is trivial and $k=2$, forming a multiplicative analogue of the twin primes conjecture. P. Berrizbeitia and Elliott [Ramanujan J. 2 (1998), no. 1-2, 219--223; MR1642879 (2000a:11122)] were able to obtain the explicit value $k=19$. The author now improves this result by showing that $k=9$ is admissible. This is derived from the following theorem: Given an arbitrary positive rational number $\gamma$, integers of the form $(p+1)/\gamma(q+1)$ with $p,q$ prime have density at least $\frac{11}{40}$. This improves a result of J. Meyer and the reviewer [Bull. Sci. Math. (2) 108 (1984), no. 4, 437--444; MR0784678 (86h:11072)], who showed the lower bound $\frac14$ when $\gamma$ is an integer. An approach to replace $\frac{11}{40}$ by $\frac{3}{10}$ is briefly outlined.

Unfortunately I am not well versed enough in multiplicative number theory to do the results justice by indicating their significance. I will just say that "product bases" results tend to be very difficult, and that the study of large sieve method is turning into an essential tool in modern analytic number theory. Conjectures such as this one, though I find them intrinsically interesting, tend to be valued because of the tools that provide us with. I expect that the reviews above indicate some of this.

[Edit (Feb. 17/2012): Numerical work related to this sequence can be found in Matthew M. Conroy, "A Sequence Related to a Conjecture of Schinzel", Journal of integer sequences, vol. 4 (1) (2001).]

  • $\begingroup$ The reference was there in the text, but i was in some other language. $\endgroup$
    – anonymous
    Jan 20, 2011 at 22:09
  • $\begingroup$ I tracked the papers down last term (and also tried unsuccessfully to find additional work on the conjecture). They are in English, but for some reason appear listed as being in German by some databases. $\endgroup$ Jan 20, 2011 at 22:14
  • $\begingroup$ I have posed this question on MO to have more views regarding this problem. $\endgroup$
    – anonymous
    Jan 29, 2011 at 19:49

That the numbers $p/q$, $p$, $q$ prime, are dense in the positive reals is a simple consequence of the Prime Number Theorem, q.v.

  • $\begingroup$ This i already knew. I have even posted that question somewhere. $\endgroup$
    – user9413
    May 9, 2011 at 11:17

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