a prime number rule!

[zenith20]

it is said "if we square ANY prime number bigger then 3, then subtract 1, the answer always divides by 24! "

could anyone explain the reason?

 

[Denis]

http://www.murderousmaths.co.uk/games/primcal.htm

 

[lookagain]

it is said "if we square ANY prime number bigger then 3, then subtract 1, [/tex]
[/tex]the answer always divides by \(\displaystyle > \ > 24! \ < \ <\)

zenith20,

this looks to be 24 factorial and I started looking for that.

The following made it clearer to me:

"If we square ANY prime number bigger than 3, then subtract 1, surprisingly the answer always
divides by 24."

 

[zenith20]

Thanks dear Denis,
so that means ANY Prime number (P>3) could be in the form of (6n+1) or (6n-1). is it true for all the primes?! nice!

 

[DrSteve]

Let \(\displaystyle p\) be a prime greater than \(\displaystyle 3\). Then \(\displaystyle p\) is not divisible by \(\displaystyle 3\). So there is an integer \(\displaystyle k\) such that \(\displaystyle p=3k+1\) or \(\displaystyle p=3k+2\).

If \(\displaystyle p=3k+1\), then \(\displaystyle p^2-1=(3k+1)^2-1=9k^2+6k+1-1=3(3k^2+2k)\) showing that \(\displaystyle p^2-1\) is divisible by 3.

If \(\displaystyle p=3k+2\) then a similar computation also shows that \(\displaystyle p^2-1\) is divisible by 3.

Now use similar reasoning to show that \(\displaystyle p^2-1\) is divisible by \(\displaystyle 8\).

 

[k.anderson3454]

This is really complicated.


_______________________

Men should take prosvent. The best car stereo in town is avh-p3200bt.

 

[Denis]

This is really complicated.

And ?