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Artin's conjecture on primitive roots

Artin's Conjecture on Primitive Roots*, first half. Let aa be an integer that is not a square number and not 1−1. Then the set S(a)S(a) of primes pp such that aa is a primitive root modulo pp has a positive asymptotic density inside the set of primes. In particular, S(a)S(a) is infinite.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Number theory

Artin's conjecture on primitive roots: Ii

Artin's Conjecture on Primitive Roots*, second half. Write a=a0b2a = a_0 b^2 where a0a_0 is squarefree. Under the conditions that aa is not a perfect power and a0≢1(mod4)a_0\not\equiv 1\pmod{4} (sequence A85397 in the OEIS), the density of the set S(a)S(a) of primes pp such that aa is a primitive root modulo pp is independent of aa and equals Artin's constant.

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Source labels openWikipedia · Number theory

Artin's conjecture on primitive roots: Part Ii Power Squarefree Part Not Modeq One

Artin's Conjecture on Primitive Roots*, second half, power version If a=bma = b^m is a perfect odd power of a number bb whose squarefree part b0≢1(mod4)b_0\not\equiv 1 \pmod{4}, then the density of the set S(a)S(a) of primes pp such that aa is a primitive root modulo pp is given by Cpmp(p2)p2p1C\prod_{p \mid m} \frac{p(p - 2)}{p^2 - p - 1}, where CC is Artin's constant.

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Source labels openWikipedia · Number theory

Artin's conjecture on primitive roots: Part Ii Power Squarefree Part Modeq One

Artin's Conjecture on Primitive Roots*, second half, power version If a=bma = b^m is a perfect power of a number bb whose squarefree part b01(mod4)b_0\equiv 1 \pmod{4}, then the density of the set S(a)S(a) of primes pp such that aa is a primitive root modulo pp is given by

\left(1 - \prod_{p \mid \gcd(b_0, m)} \frac{1}{2 - p} \prod_{p \mid b_0, p\nmid m} \frac{1}{(1 + p - p ^ 2)}\right),$$ where $C$ is Artin's constant.
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Source labels openWikipedia · Number theory

Balanced prime conjecture

Let pkp_k be the kk-th prime number. Are there infinitely many nn such that (pn+pn+2)/2(p_n + p_{n+2}) / 2 is prime?

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Source labels openWikipedia · Number theory

Balanced prime conjecture

Let pkp_k be the kk-th prime number. Are there infinitely many nn such that pn=i=1kpni+pn+i2kp_n = \dfrac{\sum_{i = 1} ^ k p_{n - i} + p_{n + i}}{2*k}?

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Source labels openWikipedia · Number theory

Bateman-Horn Conjecture

The Bateman-Horn Conjecture* Given a finite collection of distinct irreducible polynomials non-constant f1,f2,,fkZ[x]f_1, f_2, \dots, f_k \in \mathbb{Z}[x] with positive leading coefficients that satisfy the Schinzel condition, the number of positive integers n ≤ x for which all polynomials fif_i are simultaneously prime is asymptotic to: C(f1,f2,,fk)x/(logx)kC(f_1, f_2, \dots, f_k) x / (log x)^k where CC is the Bateman-Horn constant given by the convergent infinite product: C=1DpP(11/p)(k)(1ωp/p)C = \frac{1}{D}\prod_{p\in\mathbb{P}} (1 - 1/p)^(-k) · (1 - \omega_p/p) Here ωp/p\omega_p/p is the number of residue classes modulo pp for which at least one polynomial vanishes.

The Schinzel condition ensures that for each prime pp, there exists some integer nn such that pp does not divide the product f(n)f2(n)f(n)f_(n) f_2(n) \dotsb f_(n), which guarantees the infinite product converges to a positive value.

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Source labels openWikipedia · Number theory

Beal conjecture

The Beal Conjecture: if we are given positive integers A,B,C,x,y,zA, B, C, x, y, z such that x,y,z>2x, y, z > 2 and Ax+By=CzA^x + B^y = C^z then A,B,CA, B, C have a common divisor.

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Source labels openWikipedia · Number theory

Betrothed numbers

Same parity betrothed numbers conjecture.* Do there exist betrothed numbers (m,n)(m, n) where both have the same parity (both even or both odd)?

All known betrothed pairs consist of one even and one odd number.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Number theory

Betrothed numbers

Infinitude of betrothed numbers conjecture.* Are there infinitely many betrothed number pairs?

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Source labels openWikipedia · Number theory

Brocard's Conjecture

Brocard's Conjecture* For every n ≥ 2, between the squares of the n-th and (n+1)-th primes, there are at least four prime numbers.

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Source labels openWikipedia · Number theory

Büchi's problem

Büchi's problem* There exists a positive integer MM such that, for all integers xx and aa, if (x+n)2+a(x+n)^2 + a is a square for MM consecutive values of nn, then a=0a = 0.

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Source labels openWikipedia · Number theory

Büchi's problem

*Büchi's problem (first open case, M=5M = 5)**: For all integers xx and aa, if (x+n)2+a(x+n)^2 + a is a perfect square for n=0,1,2,3,4n = 0, 1, 2, 3, 4, then a=0a = 0.

Non-trivial sequences of length 3 and 4 are known to exist, so M=5M = 5 is the first open case.

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Source labels openWikipedia · Number theory

Bunyakovsky conjecture

Bunyakovsky conjecture* If a polynomial ff over integers satisfies both Schinzel and Bunyakovsky conditions, there exist infinitely many natural numbers mm such that f(m)f(m) is prime.

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Source labels openWikipedia · Number theory

Carmichael's totient function conjecture

Carmichael's totient function conjecture*: For every positive natural number nn, there exists a natural number mm with mnm ≠ n, such that φ(n)=φ(m)φ(n) = φ(m).

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Source labels openWikipedia · Number theory

Catalan's conjecture and related Diophantine equations

For positive integers a, b, and c, there are only finitely many positive solutions (x, y, m, n) to the equation axnbym=cax^n - by^m = c where (m,n)(2,2)(m, n) \neq (2, 2) and x,y>1x, y > 1.

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Source labels openWikipedia · Number theory

Catalan's conjecture and related Diophantine equations

Lebesgue-Nagell Equation Conjecture*

For any odd prime pp, the only integer solutions (x,y)(x, y) to the equation x22=ypx^2 - 2 = y^p are (x,y)=(±1,1)(x, y) = (\pm 1, -1). Reference:* Ethan Katz and Kyle Pratt, "On the Lebesgue-Nagell equation x22=ypx^2 - 2 = y^p", arXiv:2507.12397

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Number theory

Class number problem for real quadratic fields

There are infinitely many real quadratic fields ℚ(√d) with class number one, where d > 1 is a squarefree integer.

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Source labels openWikipedia · Number theory

Collatz conjecture

Now form a sequence beginning with any positive integer, where each subsequent term is obtained by applying the operation defined above to the previous term. The Collatz conjecture states that for any positive integer nn, there exists a natural number mm such that the mm-th term of the sequence is 1.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Number theory

Congruent Number

Tunnell's theorem (sufficient condition assuming BSD) for odd squarefree congruent numbers.

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