We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
118
3
avatar

Prove that for every Prime number P ≥ 5, P^2 - 1 mod 24 = 0. Thank you for help.

 Nov 28, 2018
 #1
avatar+701 
0

Something to do with the Mersenne primes \(2^n - 1\).

 

Sorry, I am not sure how to solve this. 

 Nov 29, 2018
 #2
avatar+21990 
+10

Prove that for every Prime number P ≥ 5, P^2 - 1 mod 24 = 0. Thank you for help.

 

\(\text{p in the form $4n+1$:} \begin{array}{|rcll|} \hline && p^2 - 1 \\ &=& (4n+1)^2 - 1 \\ &=& 16n^2+8n+1-1 \\ &=& 8n(n+1) \quad\Rightarrow \quad 8 | p^2-1 \\ \hline \end{array}\)

 

\(\text{p in the form $4n-1$:} \begin{array}{|rcll|} \hline && p^2 - 1 \\ &=& (4n-1)^2 - 1 \\ &=& 16n^2-8n+1-1 \\ &=& 8n(n-1)\quad \Rightarrow \quad 8 | p^2-1 \\ \hline \end{array}\)


\(\text{Euler's theorem}\\ \text{This states that if $a$ and $n$ are relatively prime $gcd(a,n) = 1$} \\ \text{then} \\ {\displaystyle a^{\varphi (n)}\equiv 1\mod n.} \)

 

\(\begin{array}{|rcll|} \hline a^{\varphi (3)} &\equiv& 1\pmod{3} \quad | \quad \varphi (3) = 2 \\ a^2 &\equiv& 1\pmod{3} \quad | \quad \text{if } gcd(a,3) = 1, \text{ so $a$ is a prime number and not } 3 \\ p^2 &\equiv& 1\pmod{3} \quad | \quad p\ne 3! \\ \text{so } p^2 -1 &\equiv& 0 \pmod{3} \quad\Rightarrow \quad 3 | p^2-1 \\ \hline \end{array}\)

 

\(\text{If $8 | p^2-1$ and $3 | p^2-1$ so also divides $24|p^2-1$}\)

 

laugh

 Nov 29, 2018
 #3
avatar
+2

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

Guest Nov 29, 2018

39 Online Users

avatar
avatar
avatar
avatar
avatar
avatar