Prove that for every Prime number P ≥ 5, P^2 - 1 mod 24 = 0. Thank you for help.
+773 Something to do with the Mersenne primes \(2^n - 1\).
Sorry, I am not sure how to solve this.
+26367 Prove that for every Prime number P ≥ 5, P^2 - 1 mod 24 = 0. Thank you for help.
\(\text{p in the form $4n+1$:} \begin{array}{|rcll|} \hline && p^2 - 1 \\ &=& (4n+1)^2 - 1 \\ &=& 16n^2+8n+1-1 \\ &=& 8n(n+1) \quad\Rightarrow \quad 8 | p^2-1 \\ \hline \end{array}\)
\(\text{p in the form $4n-1$:} \begin{array}{|rcll|} \hline && p^2 - 1 \\ &=& (4n-1)^2 - 1 \\ &=& 16n^2-8n+1-1 \\ &=& 8n(n-1)\quad \Rightarrow \quad 8 | p^2-1 \\ \hline \end{array}\)
\(\text{Euler's theorem}\\ \text{This states that if $a$ and $n$ are relatively prime $gcd(a,n) = 1$} \\ \text{then} \\ {\displaystyle a^{\varphi (n)}\equiv 1\mod n.} \)
\(\begin{array}{|rcll|} \hline a^{\varphi (3)} &\equiv& 1\pmod{3} \quad | \quad \varphi (3) = 2 \\ a^2 &\equiv& 1\pmod{3} \quad | \quad \text{if } gcd(a,3) = 1, \text{ so $a$ is a prime number and not } 3 \\ p^2 &\equiv& 1\pmod{3} \quad | \quad p\ne 3! \\ \text{so } p^2 -1 &\equiv& 0 \pmod{3} \quad\Rightarrow \quad 3 | p^2-1 \\ \hline \end{array}\)
\(\text{If $8 | p^2-1$ and $3 | p^2-1$ so also divides $24|p^2-1$}\)
Here's another way:
p2-1=(p-1)*(p+1). p is a prime number larger than 2, therefore p-1 and p+1 both have to be divisible by 2. Because they are 2 numbers divisible by 2 with a difference of 2 one of them is divisible by 4, meaning that one is divisible by at least 4 and one is divisible by at least 2- therefore (p-1)*(p+1) is divisible by at least 8. Because p is larger than 3, p is noy divisible by 3. The numbers p-1, p, p+1 are consecutive, meaning that at leadt one of them is divisible by 3. P is a prime larger than 3 therefore he is not divisible by 3 meaning that one of p-1, p+1 is divisible by 3 meaning that (p-1)*(p+1) is divisible by 3.
(p1)*(p+1) is divisible by 3 and by 8 so he is divisible by 3*8=24
