Proof of “The square of any odd integer is also odd.”

Let’s prove the statement “The square of any odd integer is also odd.”

First we restate the theorem in a way that will lead us towards the proof, by introducing a variable n into the statement of the theorem to represent “any (given) odd integer”:

Theorem: If n is odd, then n^2 is also odd.

In order to prove this, we will need the definition of what it means for a integer to be odd:

Definition: An integer n is odd if n = 2k+1 for some integer k.

Proof of theorem: Assume n is an odd integer.  Therefore, by definition, n = 2k+1 for some k.  Then

    \[n^2 = n*n = (2k+1)(2k+1) = 4k^2 + 2k + 2k + 1 = 4k^2 + 4k + 1 = 2( 2k^2 + 2k ) + 1\]

 

This shows that n^2 is odd, since n^2 = 2j + 1, where j = 2k^2 + 2k, so n^2 satisfies the definition of being odd.

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