Wall-Sun-Sun primes and Fibonacci divisibility

Definitions and mathematical statement

Define the Fibonacci sequence by $F_0=0$, $F_1=1$, and $F_{m+2}=F_{m+1}+F_m$ for $m\ge0$. For a prime $p$ and nonzero integer $a$, let $v_p(a)$ be the largest exponent $e\ge0$ such that $p^e\mid a$. The first formulation of the conjecture is \[ \forall\text{ primes }p\ \exists m\ge1\qquad v_p(F_m)=1. \] The zero term $F_0$ is excluded. The standard equivalent formulation for primes $p>5$ is \[ p^2\nmid F_{p-(p/5)}. \] Here $(p/5)$ is the Legendre symbol: it is $1$ if the nonzero residue of $p$ is a square modulo five and $-1$ otherwise. For $p>5$ the integer $p$ does divide this Fibonacci number, so the nondivisibility by $p^2$ says its valuation is exactly one. The primes two, three and five are covered by $F_3=2$, $F_4=3$, and $F_5=5$.

A prime $p>5$ violating the second formula is a Wall--Sun--Sun prime. The question is their nonexistence, not a finite search-bound claim. A counterexample must verify the squared-prime divisibility at the specified index.

Short English statement

For every prime, is some positive Fibonacci number divisible by that prime exactly once?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3390 (OPG-822), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, Wall-Sun-Sun primes and Fibonacci divisibility. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/fibonacci_divisibility.