Question

Fun modular arithmetic problem I discovered on my own @perl

Answer 1

\[\LARGE a^n \mod[a(a-1)]=a\]

Answer 2

ok

Answer 3

See, the key here is that you're almost mod a^2, but not quite. So every time you square a, you divide out almost all of it leaving just one single a left.

Answer 4

A simple example: \[\large 3^2\mod 6=9\mod6=(9-6)\mod6=3 \mod 6\]

Answer 5

the mod function in the equation editor does some really weird spacing.

Answer 6

yeah i had to read that twice :)

Answer 7

ok
3^3 mod 6 = (27 - 4*6) mod 6 = 3 mod 6

in general
a^n - k*(a(a-1)) = a mod (a(a-1)) for k big enough

Answer 8

therefore
a^n = a mod (a(a-1))

Answer 9

what was your problem that you did in the chat box

Answer 10

Well I like to think of it, as long as you have \[\LARGE a^2\mod k = a\] You have all of them.

Answer 11

right because
a^3 = a^2 * a , etc

Answer 12

Yeah, I also think it's generalizable, I haven't gotten around to it yet but like if you do something it will only equal itself for every 3rd power etc... So \[\LARGE a^{3n}\mod k= a\] That kind of thing. Or easily 3n+1 or 6n+4 or whatever you wish.

Answer 13

I suspect it's something like: \[\LARGE a^{nk} \mod (a^k-1) = a\] But I don't know if that's quite right...

Answer 14

im looking over this again

so if you start with
a mod (a^2 - a )

a^2 mod (a^2 -a ) = a because a^2 is bigger than (a^2 - a) by 'a'

Also you might have to say, this is only valid for a > 2
1^2 mod (1-1) = 0
2^2 mod (2^2 -2) = 4 mod 2 = 0

3^2 ( mod 3^2 - 3 ) = 3

and then by modular math
a^n = a^(2n)= (a^2)*(a^2) ,... = a * a ...= a
or if
a^(2n+1) = a^(2n) * a = a*a = a

Answer 15

True, I forgot about that. Although it's not at all clear what mod0 represents.

Answer 16

yeah i get an error when i try to do 1 mod 0 =
it says divide by zero , probably because the definition of mod

a mod n = remainder of (a /n ) long division style