Proving an equation holds for all integers

Question

If I'm asked to prove that an equation holds for all integers, how should I go about it?

My initial reaction would be to conduct two tests. The first one I plug in 2x into both sides of the equation and reduce and see if it comes out to the same thing on both sides. Then plug in 2x-1 into both sides and reduce and see if its the same. That would take care of both even and odd cases, but the more I think about this, the more incorrect it feels. I don't actually think that this proves anything since you are plugging the same formula into both sides. I don't know where I've seen this done before, but I know it is something that used to test something similar.

My next guess would be something like induction. Which is correct?

Answer 1

Proof by consideration of exhaustive subsets (evens and odds):

If $n$ is even, then $n\mod 2 = 0 = -n\mod 2$.

If $n$ is odd, then $n\mod 2 = 1 = -n\mod 2$.

$\blacksquare$

Answer 2

A proof by induction would be appropriate in this case. Since I don't know the specifics of the equation you are dealing with, I'll give you a general outline.

1) Let k represent the integer in your equation. 2) Show that the equation holds for k=1. 3) Then assume that it holds for k=n and then show that it holds for k=n+1.

Without knowing the specific case I'm not sure how you would show it holds for the negative counterpart of each natural number, except to simply show that the value of the function when k is negative is the negative of the value of the function when k is positive.