Could you help me prove such divisibilty, please?

Question

I'm having trouble with this:

Show that, if a>=1 is an integer and p >3 is a prime, then a^p - a is divisible by 6p, if p is even, and by 12p, if p is odd.

Thank you

Answer 1

There's a small problem with your question.
If p>3 and p is prime, then p is automatically odd!
Also note that p <>2 or 3, so (p,12) =1.
So, we have to prove 3 things:
1). a^p-a is divisible by p.
This follows from Fermat's little theorem.

2), a^p-a is divisible by 3.
Let's factor a^p-a:
We get
a^p-a = a(a^(p-1)-1) = a( a^(p-1)/2 - 1)(a^(p-1)/2 + 1).
If (p-1)/2 is a power of 2(e.g. if p is a Fermat prime)
then the second factor is divisible by a-1 and a+1.
If (p-1)/2 contains an odd prime factor
then the second factor is divisible by a-1 and
the third by a+1.
In any case a^p is divisible by a-1, a and a+1
and one of these 3 consecutive numbers is
divisible by 3.

3). a^p -a is divisible by 2 or 4.
Again, let's write it as
a(a^(p-1)-1)
Let's look at the possibilities mod 4:
a = 0. Then we certainly have divisibility by 4.
a =1. Then a^(p-1) -1 =0(mod 4), so again we have
divisibility by 4.
a = 2. Now a^(p-1)-1 is odd so we only have divisibility
by 2.
a = 3. Now a^(p-1) -1 = 0(mod 4), since p-1 is even.

So a^p -a is divisible by 12p unless a = 2(mod 4)
in which case it is divisible by 6p.

Answer 2

well by divisble if you mean tht you will only get an integer this is not true then.....and if by divisble if you mean you get a decimal as an answer then this is true in any case.