If n is odd, then n^2 - 3 is even

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Question from Tracy, a student:

Prove the statement:

For all integers n, if n is odd, then n2 - 3 is even.

Hi Tracy,

If $n$ is odd then it one more than an even integer so $n$ can be written $n = 2k + 1$ for some integer $k.$ Expand $n^2 - 3 = (2k + 1)^2 - 3.$

Penny

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