Erdős Problem #203

Definitions and mathematical statement

Shared notation.

Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

The condition $(m,6)=1$ means that $m$ is divisible by neither two nor three. The desired fixed $m$ makes $2^k3^\ell m+1$ composite for every pair $(k,\ell)\in{\mathbb{Z}}_{\geq0}^2$. Both exponents can vanish, so $m+1$ is also among the tested values. [S1] [S2]

Statement recorded in the supplied source.

Is there an integer $m\geq 1$ with $(m,6)=1$ such that none of $2^k3^\ell m+1$ are prime, for any $k,\ell\geq 0$? [S1]

Short English statement

Is there a fixed integer coprime to six such that multiplying it by any powers of two and three and then adding one never produces a prime?

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 1992 (EP-203), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath.

• S2 T. F. Bloom, Erdős Problems, Problem 203, https://www.erdosproblems.com/203. Reference from the supplied record; the live problem page was not independently checked for this compilation.