Are all Fermat Numbers square-free?

Definitions and mathematical statement

For every integer $n\ge0$, the $n$th Fermat number is $F_n=2^{2^n}+1$. A positive integer $N$ is square-free if no square of a prime divides it, equivalently if every prime in its prime factorization occurs with exponent one. The question is whether \[ \forall n\ge0\ \forall\text{ primes }p,\qquad p^2\nmid F_n. \] Here $n$ need not be prime and there is no upper bound on it. The numbers $F_n$ are odd, so only odd primes can divide them. One may equivalently ask whether $v_p(F_n)\le1$ for all $p,n$, where $v_p(N)$ is the largest integer $a\ge0$ such that $p^a\mid N$.

Square-freeness permits a Fermat number to have many distinct prime factors; it does not require the number itself to be prime. The indices $n=0,1$ give $3$ and $5$ and are included. Exhibiting many distinct factors, or showing that different Fermat numbers are coprime, does not exclude repeated factors within one $F_n$. A counterexample consists of a specific index $n$ and a prime $p$ with $2^{2^n}\equiv-1\pmod{p^2}$, certified by exact arithmetic.

Short English statement

Does every Fermat number have no repeated prime factor?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3372 (OPG-55810), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, Are all Fermat Numbers square-free?. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/are_all_fermat_numbers_square_free.