The square of a prime is one greater than a multiple of 24?

Question

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.

Answer 1

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.

Answer 2

Hint: If $p$ is a prime which is neither $2$ nor $3$, then $p=6n\pm1$ for some $n\in\mathbb N$.

Answer 3

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.

Answer 4

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 $24.$

Answer 5

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 $24h+1$