- #1
RichardParker
- 23
- 0
Homework Statement
If a [tex]\in[/tex] Z, then [tex]a^3 \equiv a (mod 3) [/tex].
2. The attempt at a solution
Proof: Suppose a [tex]\in[/tex] Z. Thus a is either odd or even.
Case 1: Let a be even. Thus a = 2k, for some [tex]k[/tex] [tex]\in[/tex] Z. So [tex] a^3 - a = 8k^3 -2k = 2(4k^3 - k) = 2(k)(2k - 1)(2k + 1)[/tex]. Notice that, for all [tex]k[/tex] [tex]\in[/tex] Z, [tex](k)(2k - 1)(2k + 1) = 3b[/tex], for some [tex]b[/tex] [tex]\in[/tex] Z. Thus [tex]a^3 - a = 2*3b[/tex]. This means [tex]3|(a^3 - a)[/tex]. Therefore [tex]a^3 \equiv a (mod 3) [/tex].
(I did not continue with the case of a being odd.)
My question is how do I prove that [tex](k)(2k - 1)(2k + 1) = 3b[/tex], for some [tex]b[/tex] [tex]\in[/tex] Z, is it enough that [tex](k)(2k - 1)(2k + 1) = 3b[/tex] is observable on any k?
Last edited: