Are There Infinitely Many Sophie Germain Primes?

Are There Infinitely Many Sophie Germain Primes?

Definitions and mathematical statement

A Sophie Germain prime is a positive prime $p$ for which $2p+1$ is also prime. The infinitude question is \[ \forall X>0\ \exists p>X:\quad p\text{ and }2p+1\text{ are prime}. \] Equivalently, the number of such $p\le x$ should tend to infinity as $x\to\infty$. No asymptotic counting formula, density constant, or effective bound for the next example is required. Primality of both linear expressions is essential; replacing one with an integer having two prime factors is only a weaker approximation to the question.

Formulation and concept references: [S1].

Short English statement

Are there arbitrarily large primes $p$ for which $2p+1$ is also prime?

Research attempt

Scope and source check (27 September 2026).

The assertion is infinitude, with both expressions prime. An asymptotic formula would suffice but is stronger than required; a theorem allowing an almost-prime second expression would be weaker. The inherited page [S1] records the definition and historical computations. Its dated record-size claims are not used as a current numerical frontier. The primary author exposition [E1] places this pair of linear polynomials within the Dickson and Bateman--Horn conjectures. Those conjectures are conditional inputs, not tools already proving this instance.

This attempt works through three concrete routes: local congruence admissibility, an exact factorial certificate, and the proposed prime-pair counting function. The local calculations are complete and give an explicit obstruction to every fixed finite sieve argument. The counting step that would force infinitely many primes remains missing.

Lemma 390.1: exact small-prime restrictions.

If $p>3$ and both $p$ and $2p+1$ are prime, then $p\equiv5\pmod6$. Indeed $p$ is odd and is not divisible by three. If $p\equiv1\pmod3$, then $2p+1$ is a multiple of three larger than three. Hence $p\equiv2\pmod3$. For $p>5$, additionally $p\not\equiv0,2\pmod5$, since these residues make $p$ or $2p+1$ divisible by five. Combining these restrictions gives \begin{equation}\label{wsref-0-61-33-39-30-3a-74-68-69-72-74-79} p\equiv11,23,29\pmod{30}. \end{equation} The exceptions $p=2,3,5$ really are Sophie Germain primes and must not be discarded by applying the large-input congruences to all inputs.

More generally, the forbidden residues for the product $n(2n+1)$ modulo a prime $q$ are zero and $-2^{-1}$ if $q$ is odd, and just zero if $q=2$. Let their number be $\rho(q)$, so \[ \rho(2)=1,\qquad \rho(q)=2\ (q>2). \] Each prime leaves at least one allowed residue. In fact $n\equiv-1\pmod q$ is always allowed, since the two expressions are then both $-1$ modulo $q$. This proves local admissibility directly, without testing only a finite initial list of primes.

A finite congruence sieve has an exact count.

For a finite set $S$ of primes put $M=\prod_{q\in S}q$. The Chinese remainder theorem gives exactly \begin{equation}\label{wsref-0-61-33-39-30-3a-63-6c-61-73-73-65-73} R_S=\prod_{q\in S}(q-\rho(q)) \end{equation} allowed classes modulo $M$. Each block of $M$ consecutive integers has exactly $R_S$ survivors. In particular the number of survivors up to $X$ is \[ X\prod_{q\in S}\left(1-\frac{\rho(q)}q\right) +O(R_S), \] with an absolute implied constant one if the error is compared class by class. For fixed $S$ this tends to infinity linearly in $X$. It counts integers whose two expressions have no prime divisor in $S$; it does not count prime pairs.

There is an equivalent inclusion--exclusion identity which shows what happens when the sieve grows. For squarefree $d\mid M$, define $\rho(d)=\prod_{q\mid d}\rho(q)$. The congruence $d\mid n(2n+1)$ has exactly $\rho(d)$ residue classes. Hence \begin{align*} \#\{1\leq n\leq X:\gcd(n(2n+1),M)=1\} &=X\prod_{q\mid M}\left(1-\frac{\rho(q)}q\right)+E,\\ |E|&\leq\sum_{d\mid M}\rho(d)=\prod_{q\mid M}(1+\rho(q)). \end{align*} For primes up to $z\geq2$, the last expression is $2\cdot3^{\pi(z)-1}$. It is harmless when $z$ is fixed. Setting $z$ near $\sqrt{2X+1}$, as would be needed to certify both inputs prime by trial division, makes this error estimate unusable as a positive lower bound. One cannot retain the fixed-sieve main term while silently discarding this growing error.

Proposition 390.2: arbitrarily large composite pairs survive every fixed sieve.

Fix $S$. Choose two different odd primes $a,b$ outside $S$. Solve the simultaneous congruences \[ n\equiv-1\pmod M,\qquad n\equiv0\pmod a, \qquad n\equiv-2^{-1}\pmod b. \] Their moduli are pairwise coprime, so the Chinese remainder theorem gives an entire progression modulo $Mab$. For sufficiently large positive members, $n>a$ and $2n+1>b$. Then $a$ is a proper divisor of $n$, $b$ is a proper divisor of $2n+1$, and neither expression has a prime factor in $S$.

Thus both expressions can be composite on infinitely many surviving inputs of any fixed local test. This does not contradict local admissibility or a conjectural prime-pair asymptotic; it refutes the inference that infinitely many survivors of finitely many congruences must include infinitely many prime pairs. Allowing more primes in the sieve changes the progression, so no one fixed progression in this construction disproves the original infinitude assertion either.

The construction is stronger than merely displaying a sporadic false positive. It supplies a uniform reason why every proposed proof stopping at a fixed finite set of excluded divisors has left an infinite uncontrolled population.

Lemma 390.3: a complete factorial test for each proposed pair.

For every positive integer $n$, \begin{equation}\label{wsref-0-61-33-39-30-3a-77-69-6c-73-6f-6e-68-61-6c-66} 2n+1\text{ is prime}\quad\Longleftrightarrow\quad (n!)^2\equiv(-1)^{n+1}\pmod{2n+1}. \end{equation} The identity \[ (2n)!=\prod_{j=1}^n j(2n+1-j) \equiv(-1)^n(n!)^2\pmod{2n+1} \] reduces this equivalence to Wilson's theorem. Here is the relevant proof, including its converse. For a prime $q$, all nonzero residues form a group. Pair each residue with its inverse; only $1$ and $-1$ are self-inverse, because $(x-1)(x+1)=0$ in the field. The product is therefore $-1$. Conversely, if an integer $m>1$ satisfies $(m-1)!\equiv-1\pmod m$ and $q<m$ is a prime divisor of $m$, then $q$ divides $(m-1)!$ but not $-1$, a contradiction. Thus $m$ must be prime.

In particular an odd prime $p$ is a Sophie Germain prime precisely when $(p!)^2\equiv1\pmod{2p+1}$. The case $p=2$ has the opposite sign in (a390:wilsonhalf), correctly giving $4\equiv-1\pmod5$. One may combine this with the separate test $(p-1)!\equiv-1\pmod p$ to certify both primalities without assuming the first.

This is a finite decision procedure, not a distribution theorem. Computing the factorial modulo $2n+1$ uses $n$ modular multiplications, so even the elementary implementation has running time exponential in the binary input length $\log n$. Faster primality algorithms would improve that cost but would still decide only specified inputs. Decidability of primality does not supply an unbounded sequence of successful inputs.

What the conditional counting law would say.

Let $A(X)=\#\{p\leq X:p,2p+1\text{ prime}\}$. The Bateman--Horn prediction for these two linear polynomials is \begin{equation}\label{wsref-0-61-33-39-30-3a-63-6f-6e-64-69-74-69-6f-6e-61-6c} A(X)\sim 2C_2\int_2^X\frac{dt}{\log t\,\log(2t+1)}, \qquad C_2=\prod_{q>2}\left(1-\frac1{(q-1)^2}\right). \end{equation} This paragraph explicitly assumes the prediction; it does not prove it. The constant follows by comparing local survival with the independent-primality model: \[ \prod_q\frac{1-\rho(q)/q}{(1-1/q)^2} =2\prod_{q>2}\frac{q(q-2)}{(q-1)^2}=2C_2. \] The product $C_2$ is positive. Each factor is positive, and the series of its deficits is bounded by $\sum_{m\geq2}m^{-2}<\infty$. For deficits at most $1/4$, the bound $-\log(1-u)\leq2u$ proves convergence of the logarithms to a finite real number.

The integral diverges: for large $X$, its contribution from $[X/2,X]$ is at least $X/(2\log X\log(2X+1))$. Thus (a390:conditional) would establish infinitude. The weaker qualitative Dickson conjecture would also suffice because the linear forms have positive leading coefficients and the verified absence of a fixed prime divisor of their product. Both routes put the missing simultaneous-primality assertion into a stated conjectural hypothesis.

A product of local densities is not automatically an asymptotic density for prime events. The events involving divisibility by different primes interact with the size restriction, and replacing all primality tests by finitely many divisibility tests leaves the false positives constructed above. Likewise a heuristic summation of prime probabilities over all $n$ is not a theorem that the deterministic sequence has infinitely many successes.

Why ordinary prime infinitude and progression theorems stop short.

Euclid's proof can force a new prime divisor of a constructed integer. It does not control whether that divisor has prime double-plus-one. Factoring a number of the form $2N+1$ also produces no reason that $(q-1)/2$ is prime for a divisor $q$. The two required expressions must be linked by the same integer $p$, not just individually appear somewhere among prime divisors.

Even Dirichlet's theorem applied to an allowed progression in (a390:thirty) proves primality of one linear expression at infinitely many inputs; it leaves the other expression uncontrolled. Intersecting two infinite sets can give a finite or empty intersection. An inference from separate prime-producing statements therefore needs a theorem about their correlation. The requested result is one such correlation statement.

A particularly precise target is simply $A(X)\to\infty$. One does not have to prove (a390:conditional), an error term, or an effective upper bound for the next prime pair. An attack that cannot achieve the full asymptotic may still succeed by proving any unbounded lower bound. None of the elementary sieve identities here supplies one for the actual prime-pair count.

Exact diagnostics and outcome.

The companion program sieves up to $200001$, records all pairs with $p\leq100000$, and checks the exceptional inputs and the congruence restrictions. It separately verifies the factorial criterion for $1\leq n\leq300$, including composite $n$ and both parity cases. It counts all residue classes for the sieve by primes through 17 and compares the count with (a390:classes). Finally it constructs explicit large composite survivors by adding the two CRT conditions with primes 19 and 23. All primality claims in these finite checks use an exact sieve, not a probable-prime test.

UNRESOLVED. The completed results are local admissibility and exact finite counts, an infinite family of false positives for every fixed sieve, and a complete one-input primality certificate. The first unproved step is a lower bound for simultaneous primality that tends to infinity while the input range grows. The conditional prime-tuple prediction identifies a sufficient statement, but no unconditional proof or counterexample to infinitude is obtained here.

Sources

• S1 Sophie Germain Prime (MathWorld). https://mathworld.wolfram.com/SophieGermainPrime.html. Formulation source.
• E1 S. L. Aletheia-Zomlefer, L. Fukshansky, and S. R. Garcia, The Bateman--Horn Conjecture: Heuristics, History, and Applications, Expositiones Mathematicae 38 $2020$, 430--479; arXiv:1807.08899v4. https://arxiv.org/abs/1807.08899. Primary author exposition checked 27 September 2026 for the prime-tuple framework. Its conjectural predictions are kept conditional.