/sci/ - Science & Math

[Anonymous No.2604725]

Hey /sci/,

So recently my number theory class has taken over my life. I need to show that n^6 - 1 is divisible by 18, where n is a prime > 3
I'm trying to show that n^6 - 1 is divisible by both 9 and 2... it's obviously divisible by 2 because it's even.
Does anyone have any idea for how I can show it's divisible by 9???

[Anonymous No.2604737]

Factor it. Since n^3 isnt divisible by 3 since n is prime, one of n^3-1 or n^3+1 must be.

Also, number theory is shit and will turn you into a soulless fuckhead.

[Anonymous No.2604741]

>>2604737

Forgot to mention you still have to come up with another factor of 3. But that's all I got without having to actually think about it, and I'm not about to do that. :P

[Anonymous No.2604754]

>>2604737
don't be dissin the Queen of Mathematics

[Anonymous No.2604791]

>>2604737
Thanks. At least I have something else to play around with now.

[Anonymous No.2604824]

>>2604791

I lied, I'm a sucker for math problems. You can probably do it by just computing n^3 mod 9 for each case for n mod 9

[Anonymous No.2605485]

>>2604824
That or just showing that n^3 mod 9 can only be 1 or 8 would work. But I can't figure out how to find n^3 mod 9 based on what n mod 9 is....

[Anonymous No.2605922]

>>2605485

If you're still here, then multiplication "carries over to mods". That is, (a mod 9)(b mod 9) = ab mod 9. So, if you know that n = 1 mod 9, then n^3 = 1^3 mod 9 = 1 mod 9. If n = 2 mod 9, then n^3 mod 9 = 8 mod 9. If n = 4 mod 9, then n^4 mod 9 = 1 mod 9, and so on. You don't need to check 3 or 6 because then n wouldn't be prime.

[Anonymous No.2605965]

(n^6-1) = (n^2-1)(n^2-n+1) (n^2+n+1)

thus it's divisible by 3