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