Primes(1)!

We note that all primes $p \ge 5$ , are odd numbers indivisible by 3. We can express any odd number larger than 3 which is indivisible by 3 as $6k \pm 1$ , where $k$ is a positive integer. Then we have:

$\begin{aligned} (6k \pm 1)^2 & = 36k^2 \pm 12k + 1 \\ & = 12k(3k \pm 1) + 1 \end{aligned}$

Since when $k$ is odd, $3k \pm 1$ is even and when $k$ is even, $3k \pm 1$ is odd, $k(3k \pm 1)$ is always even and we can express it as $2n$ , where $n$ is a positive integer. Therefore, $(6k \pm 1)^2 = 24n+1$ . Yes , the square of any prime larger than 3 can be expressed as $24n+1$ .

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$p^2 - 1 = (p - 1)(p + 1)$

Now, $(p - 1) , (p + 1)$ are two numbers that lie before and after $p$ respectively.

Since, $p$ is a prime, it has no factors other than $1$ and itself.

Since, $(p - 1) , p$ and $(p + 1)$ are three consecutive integers and $p$ is odd, both $(p - 1)$ and. $(p + 1)$ are even i.e, have $2$ as a factor.

Now, since $p \geq 5$ either of $(p - 1)$ or $(p + 1)$ have $4$ as a factor.

Since, $(p - 1) , p$ and $(p + 1)$ are three consecutive integers , one of them has $3$ as their factor. Since, $p$ is prime, the other two integers have $3$ as a factor.

Hence, we can conclude that, $(p - 1)(p + 1)$ has $2,3,4$ as its factor.

Therefore, $(p - 1)(p + 1) = (2\times 4\times 3 )n = 24\times n \Rightarrow\boxed{ p^2 = 24\times n + 1}$

Hence, $\boxed{Yes}$ it holds true for any prime greater than 5.

$Note:$ This holds true even for $2$ and $3$ . But here $n$ is not an integer.

$2^2 = 4 = 3 + 1 = 24\times \dfrac {1}{8} + 1$

$3^2 = 9 = 8 + 1 = 24\times \dfrac {1}{3} + 1$