Brocard's conjecture

Unsolved problem in mathematics

Are there at least 4 prime numbers between two consecutive squared prime numbers?

Introduction

In number theory, Brocard's conjecture is the conjecture that there are at least four prime numbers between (p_n)² and (p_n+1)², where p_n is the n^th prime number, for every n ≥ 2.^[1] The conjecture is named after Henri Brocard. It is widely believed that this conjecture is true.^[2] However, it remains unproven as of 2025. Legendre's conjecture, which states that there is a prime between consecutive integer squares, directly implies that there are at least two primes between prime squares for p_n ≥ 3 since p_n+1 − p_n ≥ 2.^[3]

Mathematical statement

Let $p_{n}$ be the $n$ -th prime, and let $\pi (x)$ be the number of prime numbers $\leq x$ . Formally, Brocard's conjecture claims:

$\pi {\big (}(p_{n+1})^{2}{\big )}-\pi {\big (}(p_{n})^{2}{\big )}\geq 4\quad {\text{for }}n\geq 2$

This is equivalent to saying that there are at least four primes between squared consecutive primes other than $2$ and $3$ .

Relation to other open problems in mathematics

Legendre's conjecture

Legendre's conjecture claims that there is a prime number between $(n)^{2}$ and $(n+1)^{2}$ for all natural number $n$ . It is an unsolved problem in mathematics as of 2025. If Legendre's conjecture is true, it immediately implies a weak version of Brocard's conjecture:^[4]

$\pi {\big (}(p_{n+1})^{2}{\big )}-\pi {\big (}(p_{n})^{2}{\big )}\geq 2\quad {\text{for }}n\geq 3$

Cramér's conjecture

Cramér's conjecture claims that $p_{n+1}-p_{n}=O((\log p_{n})^{2})$ , which gives a bound on how far apart primes can be. Cramér's conjecture implies Brocard's conjecture for sufficiently large $n$ .^[3]

Oppermann's conjecture

Oppermann's conjecture claims that there is a prime in the interval $(n^{2},n(n+1))$ and in the interval $(n(n+1),(n+1)^{2})$ . This unsolved problem directly implies Brocard's conjecture.

More information We begin with the fact that

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Proof that Oppermann's Conjecture implies Brocard's Conjecture

We begin with the fact that $p_{n+1}-p_{n}\geq 2$ , meaning that the minimal interval between primes is $(p_{n},p_{n}+2)$ . Then, according to Oppermann's conjecture, there is a prime in the interval $(p_{n}^{2},p_{n}(p_{n}+1))$ , a prime in the interval $(p_{n}(p_{n}+1),(p_{n}+1)^{2})$ , a prime in the interval $((p_{n}+1)^{2},(p_{n}+1)(p_{n}+2))$ , and a prime in the interval $((p_{n}+1)(p_{n}+2),(p_{n}+2)^{2})$ . Then, we have:

$\underbrace {{\big (}p_{n}^{2},p_{n}(p_{n}+1){\big )}} _{{\text{At least }}1{\text{ prime}}},\quad \underbrace {{\big (}p_{n}(p_{n}+1),(p_{n}+1)^{2}{\big )}} _{{\text{At least }}1{\text{ prime}}},\quad \underbrace {{\big (}(p_{n}+1)^{2},(p_{n}+1)(p_{n}+2){\big )}} _{{\text{At least }}1{\text{ prime}}},\quad \underbrace {{\big (}(p_{n}+1)(p_{n}+2),(p_{n}+2)^{2}{\big )}} _{{\text{At least }}1{\text{ prime}}}$

Which implies at least $4$ primes between $p_{n}^{2}$ and $(p_{n}+2)^{2}$ , and because $p_{n}+2\leq p_{n+1}$ , there are at least $4$ primes between any two squared consecutive primes, which is exactly what Brocard's conjecture claims.

Close

Examples

More information

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n	$p_{n}$	$p_{n}^{2}$	Prime numbers	$\Delta$
1	2	4	5, 7	2
2	3	9	11, 13, 17, 19, 23	5
3	5	25	29, 31, 37, 41, 43, 47	6
4	7	49	53, 59, 61, 67, 71, ...	15
5	11	121	127, 131, 137, 139, 149, ...	9
$\Delta$ stands for $\pi (p_{n+1}^{2})-\pi (p_{n}^{2})$ .

Close

It is easy to verify the conjecture for small $n$ :

$\pi {\big (}5^{2}{\big )}-\pi {\big (}3^{2}{\big )}=9-4\geq 4,\quad \pi {\big (}7^{2}{\big )}-\pi {\big (}5^{2}{\big )}=15-9\geq 4$

The number of primes between prime squares is 2, 5, 6, 15, 9, 22, 11, 27, ... OEIS: A050216. See the table (right) for a list of primes sorted by the difference. See the animation (right) for the first 30 differences.

Current research and results

Unconditional results

Bertrand's postulate

A trivial result from Bertrand's postulate, a proven theorem, states that because there is a prime in the interval $(n,2n)$ , and the length of the interval $((p_{n})^{2},(p_{n+1})^{2})$ is much greater than $(p_{n},2p_{n})$ , Bertrand's postulate suggests many primes in the interval $((p_{n})^{2},(p_{n+1})^{2})$ , though not a sharp bound.

Baker-Harman-Pintz bound

Using the bound proven by Baker et al.,^[5] that $p_{n+1}-p_{n}<p_{n}^{0.525}$ , one can show that there exist infinitely many $p_{n}$ such that there is at least one prime in the interval $((p_{n})^{2},(p_{n+1})^{2})$ , which is a much weaker result than Brocard's conjecture.

Conditional results

Legendre's Conjecture - weak version of Brocard's conjecture

As shown above, Legendre's conjecture implies a weak version of Brocard's conjecture but is a strictly weaker conjecture.

Oppermann's Conjecture - full proof of Brocard's conjecture

As shown above, Oppermann's conjecture directly implies Brocard's conjecture for large enough $n$ , which constitutes a proof of Brocard's conjecture.

Cramér's Conjecture - full proof of Brocard's conjecture

As shown above, Cramér's conjecture implies Brocard's conjecture directly.

The Riemann Hypothesis - full proof of Brocard's conjecture

The Riemann Hypothesis implies the bound $p_{n+1}-p_{n}=O{\big (}{\sqrt {p_{n}}}\log(p_{n}){\big )}$ , which implies Brocard's conjecture for sufficiently large $n$ , similarly to Cramér's conjecture.^[6]