Is ${F_{n}}^2 - 28$ always a composite number? The problem is as follows:

Prove or disprove that if ${F_{n}}$ is $n$-th Fibonacci number, and $n>5$, then $${F_{n}}^2 - 28$$
cannot be a prime.

I came across this problem accidentally while trying to solve another problem.
I suspect that there is an identity involving ${F_{n}}^2 - 28$ that proves that it is a composite. On the other hand, it is a kind of odd and counter-intuitive to imagine an identity involving number $28$...
 A: Edit:
I'll keep my original at the bottom since it has some value under this question.
Without considering $n:F_n^2-28\lt 0$ (for example, $n=4,F_4^2-28=9-28=-19$ or similarly for $n=5$), the first couple values certainly are composite, but not in an immediately-obvious manner.
Here is an answer looking at the Fermat numbers, which are not under discussion for this question but are mildly interesting in this light anyway:
Disregarding $F_0=2^{2^0}+1$, we have $3\nmid 2^{2^n}+1$ for all positive integers $n$, making $F_n^2\equiv 1\pmod 3$, and as $28\equiv 1\pmod 3$ we have $F_n^2-28\equiv 0\pmod 3$ for all $n\gt 0$.
A: You're correct: $F_n^2-28$ is never a prime for $n\geq 6$, i.e., all $n$ where $F_n^2-28\gt 0$; arguably it's prime for $n=4$ ($F_4^2-28=-19$) and $n=5$ ($F_5^2-28=-3$).
First of all, note that if $3\not\mid F_n$, then $F_n^2\equiv 1\equiv 28\pmod 3$, so $3\mid F_n^2-28$.  We can therefore assume in what follows that $3\mid F_n$ and therefore that $4\mid n$.
Now, (defining $G_n=F_n^2-28$ for convenience in what follows:) numeric evidence suggests the conjecture that for all $n$ divisible by $4$ there's some $a$ with $G_n=5a^2+24a = a\times(5a+24)$.  We can prove this (and thus provide a factorization for all $n$) as follows:
Suppose that $G_n=5a^2+24a$; in other words, $5a^2+24a-G_n=0$.  The (positive) solution to this quadratic equation is $a=\frac1{10}\left(-24+\sqrt{24^2+20G_n}\right)$.  We'll first show that this quantity is rational, and then that it's integral.  Note that $24^2+20G_n=24^2+20(F_n^2-28)=16+20F_n^2$, so $\sqrt{24^2+20G_n}=2\sqrt{4+5F_n^2}$.  But since $n$ is even, we have $4+5F_n^2=L_n^2$ where $L_n$ is the $n$th Lucas number (a sort of conjugate to the Fibonacci numbers, satisfying the same recurrence).  Clearly $2L_n-24$ is divisible by $2$; its divisibility by $5$ is equivalent to saying that $2L_n\equiv -1\bmod 5$ or that $L_n=2\bmod 5$.  But this follows since, as noted at the start, we're in the case $4\mid n$ (and because the period of the Lucas numbers mod $5$ is $4$).  This implies the integrality of $a$, which in turn implies the desired factorization of $G_n$.
A: The recurrence for the squares of the Fibonacci numbers is
$$
x_n=2x_{n-1}+2x_{n-2}-x_{n-3}
$$
Thus a repeat of a subsequence of length $3$ implies a repeat of the entire sequence.

mod $2$, we get
$$
\overbrace{0,1,1},\overbrace{0,1,1},\dots
$$
Thus, for $n\equiv0\pmod3$, we have that $F_n^2-28\equiv0\pmod2$

mod $3$, we get
$$
\overbrace{0,1,1},1,\overbrace{0,1,1},\dots
$$
Thus, for $n\not\equiv0\pmod4$, we have $F_n^2-28\equiv0\pmod3$

mod $7$, we get
$$
\overbrace{0,1,1},4,2,4,1,1,\overbrace{0,1,1},\dots
$$
Thus, for $n\equiv0\pmod8$, we have that $F_n^2-28\equiv0\pmod7$

The only thing not covered above is $n\equiv4\pmod{24}$ and $n\equiv20\pmod{24}$.
Recall that $F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$ where $\alpha$ and $\beta$ are the roots of $x^2-x-1=0$. Note that $(\alpha-\beta)^2=5$ and $\alpha\beta=-1$. Using these relations, it is not difficult to obtain
$$
\begin{align}
5(F_{2k}^2-28)
&=5F_{2k}^2-140\\
&=\alpha^{4k}+\beta^{4k}-142\\
&=\left(\alpha^{2k}+\beta^{2k}\right)^2-144\\
&=\left(L_{2k}-12\right)\left(L_{2k}+12\right)
\end{align}
$$
where $L_n=\alpha^n+\beta^n$ is a Lucas Number. In an argument similar to those above, it can be shown that $L_{2k}-12\equiv0\pmod5$ for even $k$ and $L_{2k}+12\equiv0\pmod5$ for odd $k$.
This covers the case of all even values of $n$ for $n\ge6$ since $L_6=18$ and $L_n$ is monotonically increasing after that.

The cases above cover all $n\ge6$, which are the cases for which $F_n^2-28\ge0$.  Specifically:
If $n$ is odd, $F_n^2-28=3\cdot\frac{F_n^2-28}3$
If $n=0\pmod4$, $F_n^2-28=\frac{L_n-12}{5}(L_n+12)$
If $n=2\pmod4$, $F_n^2-28=9\cdot\frac{L_n-12}3\frac{L_n+12}{15}$
The argument above can be simplified by ignoring the cases mod $2$ and mod $7$ since only the case mod $3$ and the Lucas number argument is needed.
I see that the Lucas number identity has already been noted in another answer while I've worked on this answer, but I will include it for completeness.
