Problem B. 5490. (November 2025)
B. 5490. Prove that there exist infinitely many positive integers \(\displaystyle n\) such that numbers \(\displaystyle {2^n-2025}\), \(\displaystyle {2^n-2024}\), \(\displaystyle \dots\), \(\displaystyle {2^n+2025}\) are all composite.
Proposed by Csaba Sándor, Budapest
(5 pont)
Deadline expired on December 10, 2025.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Minden \(\displaystyle -2025\leq k\leq 2025\)-re legyen \(\displaystyle p_k\) a \(\displaystyle 2^{11}+k\) egyik prímosztója (ilyen létezik, hiszen \(\displaystyle 2^{11}+k\geq 2048-2025>1\)). Legyen \(\displaystyle P=\prod_{k=-2025}^{2025}(p_k-1)\). Ekkor bármely \(\displaystyle m\geq1\) pozitív egészre
$$\begin{align*} 2^{11+mP}+k&\equiv 2^{11}\cdot(2^{p_k-1})^{mP/(p_k-1)}+k\equiv 2^{11}+k\equiv 0\quad\mod p_k\\ &\Rightarrow p_k\mid 2^{11+mP}+k, \end{align*}$$valamint \(\displaystyle 2^{11+mP}+k>2^{11}+k\geq p_k\) teljesül minden \(\displaystyle -2025\leq k\leq 2025\)-ra, azaz \(\displaystyle 2^{11+mP}+k\) minden ilyen \(\displaystyle k\)-ra összetett. Így \(\displaystyle n=11+mP\) minden \(\displaystyle m=1,2,\dots\)-ra megfelel.
Statistics:
52 students sent a solution. 5 points: Ali Richárd, Beinschroth Máté, Bodor Ádám, Bodor Noémi, Diaconescu Tashi, Ercse Ferenc, Gál Mózes, Hajba Milán, Holló Martin, Kerekes András, Körmöndi Márk, Li Mingdao, Papp Mátyás, Péter Hanna, Rajtik Sándor Barnabás, Sajter Klaus, Sánta Gergely Péter, Sarusi-Kis Balázs, Schmidt Botond, Szabó 926 Bálint, Takács András, Tóth László Pál, Vályi Nagy Ádám András, Varga 511 Vivien, Weng Chenxin, Wiener Marcell. 4 points: Balla Ignác , Baran Júlia, Terjék Temes, Vincze Marcell, Zhai Yu Fan. 3 points: 2 students. 2 points: 3 students. 1 point: 12 students. 0 point: 2 students. Unfair, not evaluated: 1 solutions.
Problems in Mathematics of KöMaL, November 2025