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Claim. Problem 4 of ELMO 2015, the 17th Ex-Lincoln Math Olympiad, in its official solutions (linked above, a four-page file created 26 June 2015 in US Mountain time, 27 June 2015 UTC, the date this page carries). Jack Gurev proposed the problem and Sam Korsky gave the official solution. For every integer $a>1$, some $2^{2^n}+a$ with $n\ge0$ is composite. The proof: put $m=v_2(a-1)$. If $p=2^{2^m}+a$ is prime, then $v_2(p-1)=m$, so $(p-1)/2^m$ is odd. Then $n=m+\varphi((p-1)/2^m)$ gives $p\mid2^{2^n}-2^{2^m}$, so $p$ divides the larger number $2^{2^n}+a$. For even $a$ we have $m=0$, and $2+a$ is even and greater than $2$. This is the order argument of Lemma 3.3 of Barschkis's note, recorded on its claim page. With $a=n$ it answers question (iii.a) of Problem 1209 no for every shift $n\ge2$. The thread comment of 13 September 2026 by the solution's author links the file and reports the olympiad provenance.

Covers. Question (iii.a) for every shift $n\ge2$. This source does not cover the shift $0$ (parity), the shift $1$ (Euler's $641\mid2^{32}+1$), negative shifts ($n+2\le1$ at $k=0$), or questions (i), (ii) and (iii.b) to (iii.d).

Standing. Pending. An olympiad's official solutions are not a journal publication and name no outside reviewer. The site credits (iii.a) to the note's author and GPT and does not mention this proof. Nothing here is this project's own review.

Depends on. No page of this wiki.