For a prime $p\ge 17$ is $\dfrac{p^2-1}{24}$ ever a prime?

Question

It was indicated in the comments of this MO question that if $p\ge5$ is a prime then $24|p^2-1$. Indeed $p=6k\pm1$ and $p^2-1=36k^2\pm12k+1-1=12k(3k\pm1)$ and exactly one of $k$ and $3k\pm1$ is even.

Let $Q(p)=\dfrac{p^2-1}{24}$ (where $p\ge5$ is a prime).
Note that $Q(5)=1$, $Q(7)=2$, $Q(11)=5$, $Q(13)=7$ are all primes
(except $1$, which used to be a prime).

On the other hand $Q(17)=12$, $Q(19)=15$, $Q(23)=22$, $Q(29)=35$, $Q(31)=40$, $Q(37)=57$, $Q(41)=70$, $Q(43)=77$, $Q(47)=92$, $Q(53)=117$, are all composite numbers, though they do not have a common factor, and I do not see an obvious pattern.

Question. Is it true that $Q(p)=\dfrac{p^2-1}{24}$ is composite whenever $p\ge17$ is a prime?
If so, how to prove this? Is it known, any references?
(Is it related in any way to quadratic residues, am I supposed to know it? :)

I ran reduce computer algebra and it tells me the answer is yes for at least all primes $p\le21331777$ (when I interrupted it). Thank you!

Answer 1

When $p-1>24$, then $24$ can't cancel any of the factors $p-1$ and $p+1$ properly, so the number $\frac{p^2-1}{24}$ is always composite for $p-1>24$, other cases can be calculate manually...

Answer 2

One of $p-1$ or $p+1$ is divisible by $4$, say $p-1$. The other one, $p+1$, is divisible by $2$. So

$$\frac{p^2-1}{24} = \frac{1}{3}\frac{p+1}{4}\frac{p-1}{2}.$$

Now, since we know the result is an integer, it must be that $3$ divides one of $\frac{p+1}{4}$ or $\frac{p-1}{2}$. So one of $\frac{1}{3}\frac{p+1}{4}$ or $\frac{1}{3}\frac{p-1}{2}$ is an integer. If $\frac{p^2-1}{24}$ is to be prime, what condition is imposed on that integer?

Answer 3

Consider all cases modulo $12$. For example, if $p\equiv1\pmod{12}$, then $p-1$ is a multiple of $12$ and $p+1$ is a multiple of $2$. In addition, since $p\ge17$, $p-1>12$ and $p+1>2$. So $p-1$ can be written as $12k_1,k_1>1$ and $p+1=2k_2,k_2>1$. So $\frac{(p-1)(p+1)}{24}=k_1k_2$. Since neither $k$'s are equal to $\pm1$, the result cannot be prime.

The remaining 3 cases follow similiarly.