Let $F_n$ be a Fibonacci number and $p$ a prime. Verify that for $p \le 61$, if $p\equiv\pm1 \pmod{5}$ then $p\mid F_{p-1}$ Define the Fibonacci entry point of $p$ to be the least integer $n$ such that $p\mid F_n$
So for example, for $p = 3$ - the Fibonacci entry point is $n = 4$ since $F_4 = 3$ and obviously $3\mid 3$.
We are also given the following previously proven statement to be used: $F_{n+i}\equiv kF_{i}\pmod{d}$
Where $d$ is an integer, s.t. $d|F_n$ and $k = F_{n+1}$
Here is what I have so far:
I start with $p\equiv -1 \pmod5$
The only prime numbers $\le 61$ such that $p\equiv -1 \pmod5$ are $19,29,59$.
So verify that it holds for $p = 19$
$19 \mid F_{18} = 2584$
Then, from the statement $F_{n+i}\equiv kF_{i}\pmod d$, letting $i = 10$
$F_{28} \equiv F_{19}F_{10} \pmod{19}$
Now, I am kind of stuck figuring out how I can show that $29 \mid F_{28}$.
Any help would be greatly appreciated!
 A: For posterity, here is a sketch of the proof of a more general statement: If $p>5$ is prime then $p|F_{p\pm 1}$ for some choice of $+$ or $-$. We shall use algebraic number theory to deal with two cases.
Case 1: $5$ is a quadratic residue modulo $p$. In this case, $p$ splits in $\mathbf{Z}\left[\frac{1+\sqrt{5}}{2}\right]=\mathbf{Z}[\varphi]$. Thus, we can write $p=\pm\pi\bar\pi$, where $\pi$ and $\bar\pi$ are conjugate primes in $\mathbf{Z}[\varphi]$ that do not differ by a unit. Write $\pi=x+y\varphi$, so $x+y\varphi\equiv 0\pmod{\pi}$. Now, if $p|y$, then $\pi|y,x$, contradiction, so $p\nmid y$. Thus, $y$ has an inverse modulo $p$, say $y'$. Then we have $\pi|p|yy'-1$, so $\varphi\equiv -xy'\pmod{\pi}$. Summarizing, $\varphi\equiv k\pmod{\pi}$ for some integer $k\not\equiv 0\pmod{p}$. By FLT, $k^{p-1}\equiv 1\pmod{p}$, so $\varphi^{p-1}\equiv k^{p-1}\pmod{\pi}$. Thus $\varphi^{p-1}\equiv 1\pmod{\pi}$. Similarly, we see that $\bar\varphi^{p-1}\equiv 1\pmod{\pi}$, so $F_{p-1}\sqrt{5}\equiv 0\pmod{\pi}$. Since $p$ and $5$ are necessarily relatively prime, $\pi|F_{p-1}$, and $\bar\pi|F_{p-1}$. Hence $\pi\bar\pi = p|F_{p-1}$ in this case.
Case 2: $5$ is a quadratic non-residue modulo $p$. We have $5^{(p-1)/2}\equiv -1\pmod{p}$, by Euler's Criterion. Now, applying the Binomial-theorem to Binet's formula yields several terms containing $\binom{p+1}{k}\equiv 0\pmod{p}$. After reducing modulo $p$ we will be left with $\dfrac{\sqrt{5}^{p+1}+1}{2^t}$ for some $t$, which is also divisible by $p$ (by working in $\mathbf{Z}[\varphi]$), so $p|F_{p+1}$ in this case.
