The square of a prime is one greater than a multiple of 24? I read that the square of any prime number, excluding 2 and 3, is one greater than a multiple of 24. Is this a conjecture or a theorem? It's hard for me to imagine how such a thing could be proven.
 A: That the number is prime is not relevant.
What matters is that the number is odd and is not a multiple of $3$.
Let $n$ be not multiple of $3$.  Then $n = 3m \pm 1$ for some $m$.
Let $n$ also be odd.  Then $3m$ is even and $m$ is even.  So $m = 2k$ for some $k$.
So $n = 3*2k \pm 1  = 6k \pm 1$.
And $n^2 = 36k^2 \pm 12k + 1$
And $36k^2 \pm 12k + 1 = 12(3k \pm 1)k + 1$.
If $k$ is even. Then $12k$ is divisible by $24$ and $n^2$ is one more than a multiple of $24$.
If $k$ is odd, then $3k^2 \pm 1$ is even.  And then $12(3k^2 \pm 1)$ is divisible by $24$ and $n^2$ is one more than a multiple of $24$.
Hence proved.
A: All positive integer are of form $$6n,6n+1,6n+2,6n+3,6n+4,6n-1$$for some $n$ of which $6n,6n+2,6n+3,6n+4$ are all composite (excluding 1,2,3) therefore any prime is of form $6n\pm 1$ leading that the square is of form $$(6n\pm 1)^2=36n^2\pm12n+1=12k+1$$. Similarly any prime is of form $4n\pm 1$ leading that the square is of form $$(4n\pm 1)^2=16n^2\pm8n+1=8k'+1$$integrating the results any such prime square is of form $24h+1$
A: Another hint:


*

*Any prime $>3$ is congruent to $1$ or $-1\bmod 3$, so its square in congruent to $1\bmod 3$.

*Any  odd prime is congruent to $\pm 1$ or $\pm 3\bmod 8$, so its square is also congruent to $1\bmod 8$.


Now, apply the  Chinese remainder theorem.
A: Hint: If $p$ is a prime which is neither $2$ nor $3$, then $p=6n\pm1$ for some $n\in\mathbb N$.
A: You want to show that $$p^2-1 =24k$$ for some integer k.
$$ p^2-1 = (p-1)(p+1)$$
since $p$ is a prime different from $2$ and $3$, $p$ is odd therefore both $p-1$ and $p+1$ are even and one of them is a multiple of $4$.
That is the product $(p-1)(p+1)$ is a multiple of $8$. 
On the other hand $P=3k+1$ or $p=3k-1$ and in either case the product $(p-1)(p+1)$ is a multiple of $3$.
That makes the product a multiple of $24.$    
