Why Does \( a^3 \equiv a \mod 3 \) Hold for All Integers \( a \)?

[RichardParker]

Homework Statement

If a \in Z, then a^3 \equiv a (mod 3).

2. The attempt at a solution

Proof: Suppose a \in Z. Thus a is either odd or even.

Case 1: Let a be even. Thus a = 2k, for some k \in Z. So a^3 - a = 8k^3 -2k = 2(4k^3 - k) = 2(k)(2k - 1)(2k + 1). Notice that, for all k \in Z, (k)(2k - 1)(2k + 1) = 3b, for some b \in Z. Thus a^3 - a = 2*3b. This means 3|(a^3 - a). Therefore a^3 \equiv a (mod 3).

(I did not continue with the case of a being odd.)

My question is how do I prove that (k)(2k - 1)(2k + 1) = 3b, for some b \in Z, is it enough that (k)(2k - 1)(2k + 1) = 3b is observable on any k?

 

[LCKurtz]

Try your argument on a³ - a without assuming a even or odd.

 

[RichardParker]

Thanks for your reply.

Without assuming a even or odd,

a(a-1)(a+1) = 3b, for some b element of the set of integers. It is observable that a^3 - a is divisible by 3 for any value of a that I plug in. But how do I prove that this will hold for all a element of integers?

Thanks once again!

 

[LCKurtz]

Thanks for your reply.

Without assuming a even or odd,

a(a-1)(a+1) = 3b, for some b element of the set of integers. It is observable that a^3 - a is divisible by 3 for any value of a that I plug in. But how do I prove that this will hold for all a element of integers?

Thanks once again!

Well, a-1, a, a+1 are three consecutive integers...

 

[sutupidmath]

Proof: Fermat's Little Theorem! Q.E.D