I have a question about prime numbers and I have no idea how to start.
Let $p$ be a prime number, $ p \neq 2 $ and $ p \neq 3 $. Prove that there is $ k \in N $ so that $ p^{2} = 24k + 1 $
Can anyone help me at least get started?
Thanks.
I have a question about prime numbers and I have no idea how to start.
Let $p$ be a prime number, $ p \neq 2 $ and $ p \neq 3 $. Prove that there is $ k \in N $ so that $ p^{2} = 24k + 1 $
Can anyone help me at least get started?
Thanks.
Hint: prove that $p^2-1=(p-1)(p+1)$ is divisible by $24$
hint
For $ p\in\{5,7,11\} $, it is easy to check your claim is true.
Assume that $ p\ge 13 $.
there are at most four cases
$$ p=12k+1 \implies p^2-1=12k(12k+2)$$ $$=\color{red}{24}k(6k+1)$$
$$p=12k+5\implies p^2-1=6(2k+1)(12k+4)$$ $$=\color{red}{24}(2k+1)(3k+1)$$
$$p=12k+7\implies p^2-1=4(3k+2)(12k+6)$$ $$=\color{red}{24}(3k+2)(2k+1)$$
$$p=12k+11\implies p^2-1=12(k+1)(12k+10)$$ $$=\color{red}{24}(k+1)(6k+5)$$