Brazilian Primes Which Are Also Sophie Germain Primes

TL;DR

This paper disproves Schott's conjecture by identifying Brazilian primes that are also Sophie Germain primes, providing a comprehensive enumeration of all such counterexamples up to 10^46.

Contribution

It is the first to explicitly find and list all Brazilian primes that are also Sophie Germain primes up to 10^46, disproving the previous conjecture.

Findings

Counterexamples up to 10^46 identified

Schott's conjecture disproved

Complete enumeration of such primes provided

Abstract

We disprove a conjecture of Schott that no Brazilian primes are Sophie Germain primes. We enumerate all counterexamples up to $10^{46}$.

Equations

$$p=b^{q-1}+b^{q-2}+\cdots +b+1\equiv 1(\mathrm{mod}3),$$

$$14781835607449391161742645225951=1309^{10}+1309^9+\cdots +1309+1.$$

Full text

**BRAZILIAN PRIMES WHICH ARE ALSO SOPHIE GERMAIN PRIMES

Jon Grantham

**Institute for Defense Analyses/Center for Computing Sciences, Bowie, Maryland

grantham@super.org

**Hester Graves

**Institute for Defense Analyses/Center for Computing Sciences, Bowie, Maryland

hkgrave@super.org

Abstract

We disprove a conjecture of Schott that no Brazilian prime is a Sophie Germain prime. We compute all counterexamples up to $10^{46}$. We prove conditional asymptotics for the number of Brazilian Sophie Germain primes up to $x$.

1 Introduction

The term “Brazilian numbers” comes from the 1994 Iberoamerican Mathematical Olympiad [8] in Fortaleza, Brazil, in a problem proposed by the Mexican math team.111The term appears as “sensato” in the original problem [6]. The authors are puzzled by the discrepancy with [8].. They became a topic of lively discussion on the forum mathematiques.net. Bernard Schott [5] summarized the results in the standard reference on Brazilian numbers.

Definition 1**.**

A Brazilian number $n$ is an integer whose base-$b$ representation has all the same digits for some $1<b<n-1$.**

In other words, $n$ is Brazilian if and only if $n=m\left({\frac{b^{q}-1}{b-1}}\right)=mb^{q-1}+\cdots+mb+m$, with $q\geq 2$. These numbers are A125134 in the Online Encylopedia of Integer Sequences (OEIS).

A Brazilian prime (or “prime repunit”) is a Brazilian number that is prime; by necessity, $m=1$ and $q\geq 3$. See A085104 in the OEIS for the sequence of Brazilian primes. In 2010, Schott [5] conjectured that no Brazilian prime is also a Sophie Germain prime.

Sophie Germain discovered her eponymous primes while trying to prove Fermat’s Last Theorem; her work was one of the first major steps towards a proof.

Definition 2**.**

A Sophie Germain prime is a prime $p$ such that $2p+1$ is also prime.**

Germain showed that if $p$ is such a prime, then there are no non-zero integers $x,y,z$, not divisible by $p$, such that $x^{p}+y^{p}=z^{p}$. If $p$ is a Sophie Germain prime, then we say that $2p+1$ is a safe prime.

It is conjectured that there are infinitely many Sophie Germain primes, but the claim is still unproven. The Bateman-Horn conjecture [1] implies that the number of Sophie Germain primes less than $x$ is asymptotic to $2C\frac{x}{\log^{2}x}$, where $C=\displaystyle\prod_{p>2}\frac{p(p-2)}{(p-1)^{2}}\approx 0.660161$.

See [3] for further information about Sophie Germain primes.

2 Enumerating Counterexamples

To aid our search, we use a few lemmas.

Lemma 1**.**

If $p={\frac{b^{q}-1}{b-1}}$ is a Brazilian prime, then $q$ is an odd prime.

Proof.

Recall $x^{q}-1$ is divisible by the $m$th cyclotomic polynomial $\Phi_{m}(x)$ for $m|q$; therefore $p$ can only be prime if $q$ is also prime. Note that $q>2$ because $b<p-1$, so $q$ is an odd prime. ∎

The preceding lemma is also Corollary 4.1 of Schott [5].

Lemma 2**.**

If $p$ is a Brazilian prime and a Sophie Germain prime, then $p\equiv q\equiv 2$ (mod $3$) and $b\equiv 1$ (mod $3$).

Proof.

If $p$ is a Sophie Germain prime, then $3$ cannot divide the safe prime $2p+1$, so $p$ cannot be congruent to $1$ (mod $3$). The number $3$ is not Brazilian, so $p\neq 3$ and thus $p\equiv 2$ (mod ${3}$).

If $3|b$, then

[TABLE]

which is a contradiction. Lemma 1 states that $q$ is an odd prime, so if $b\equiv 2$ (mod ${3})$, then $p\equiv 1$ (mod ${3}$), a contradiction. We conclude that $b\equiv 1$ (mod ${3}$), so that $q\equiv p$ (mod ${3}$), and therefore $q\equiv 2$ (mod ${3}$). ∎

For $q=5$, we use a modification of the technique described in [7] to compute a table of length-$5$ Brazilian primes up to $10^{46}$. We will describe this computation in full in a forthcoming paper [2]. Of these, $104890280$ are Sophie Germain primes. The smallest is $28792661=73^{4}+73^{3}+73^{2}+73+1$. We very easily prove the primality of Sophie German primes with the Pocklington-Lehmer test.

For $q\geq 11$, we very quickly enumerate all possibilities for $b\leq 10^{46/(q-1)}$. We find $22$ Brazilian Sophie Germain primes for $q=11$, and none for larger $q$. (We have $q<\log_{2}{10^{46}}+1<154$.) The smallest is

[TABLE]

While we disprove Schott’s conjecture, we do have a related proposition.

Proposition 1**.**

The only Brazilian prime which is a safe prime is $7$.

Proof.

If $p=b^{q-1}+\cdots+b+1$ is a safe prime, then $\frac{p-1}{2}=\frac{1}{2}(b^{q-1}+\cdots+b)$ must also be prime. This expression, however, is divisible by $\frac{b(b+1)}{2}$, which is only prime when $b=2$ and $p=7$. ∎

The list of Brazilian Sophie Germain primes is A306845 in the OEIS. The first few counterexamples were also discovered by Giovanni Resta and Michel Marcus; see the comments for A085104.

3 Conditional Results

The infinitude of Brazilian Sophie Germain primes, as well as the asymptotic number of them, is the consequence of well-known conjectures.

Proposition 2**.**

Assuming Schnizel’s Hypothesis H, there are infinitely many Brazilian Sophie Germain primes.

Proof.

Recall that Hypothesis H [4] says that any set of polynomials, whose product is not identically zero modulo any prime, is simultaneously prime infinitely often. Take our two polynomials to be $f_{0}(x)=x^{4}+x^{3}+x^{2}+x+1$ and $f_{1}(x)=2x^{4}+2x^{3}+2x^{2}+2x+3$. Then $f_{0}(b)$ is Brazilian and $f_{1}(b)=2f_{0}(b)+1$. Rather than checking congruences, it suffices to note the existence of the above primes of this form to see that the conditions of Hypothesis H are satisfied. ∎

The Bateman-Horn Conjecture [1] implies a more precise statement about the number of Brazilian Sophie Germain primes.

Proposition 3**.**

For an odd prime $q$, let $\Phi_{q}(x)$ be the $q$th cyclotomic polynomial. Assuming the Bateman-Horn Conjecture, the number of values of $b<x$ such that $\Phi_{q}(b)$ and $2\Phi_{q}(b)+1$ are simultaneously prime is [math] or $\sim C_{q}\frac{x}{{\log^{2}x}}$, for some positive constant $C_{q}$, depending on whether $\Phi_{q}(b)(2\Phi_{q}(b)+1)$ is identically zero modulo some prime $p$.

Proof.

This follows immediately from the Bateman-Horn conjecture, with $C_{q}=\left(\prod_{p}\frac{1-N(p)/p}{(1-1/p)^{2}}\right)/q^{2}$, where $N(p)$ is the number of roots of $\Phi_{q}(b)(2\Phi_{q}(b)+1)$ modulo $p$. ∎

Corollary 1**.**

Assuming the Bateman-Horn Conjecture, the number of Brazilian Sophie Germain primes up to $x$ is $\sim C\frac{x^{1/4}}{{\log^{2}x}}$, for some $C$.

Proof.

To find the number of Brazilian Sophie Germain primes less than $y$ of the form $\Phi_{q}(b)$ for a fixed $q$, we apply the preceding proposition, substituting $x=y^{1/(q-1)}$, and get $\sim C^{\prime}_{q}\frac{y^{1/{(q-1)}}}{{\log^{2}y}}$, with $C^{\prime}_{q}=C_{q}(q-1)^{2}$. We sum over all $q\equiv 2$ (mod $3$) and notice that the $q=5$ term dominates. We can thus take $C=C^{\prime}_{5}$. ∎

**Acknowledgement. ** Thanks to Enrique Treviño for helping track down the reference in [6], and to the referee for helpful comments.