Prove if n2n^2 is even, then nn is even.

I am just learning maths, and would like someone to verify my proof.

Suppose n is an integer, and that n2 is even. If we add n to n2, we have n2+n=n(n+1), and it follows that n(n+1) is even. Since n2 is even, n is even.

Is this valid?


It could use a little more explanation, but yes, it works. I’d expand it to point out explicitly why n(n+1) is even and that n=(n2+n)n2 is then the difference of two even numbers and as such is even (assuming that you already have this fact available to use).

An alternative approach is to show that if n is odd, then n2 is odd; the desired result is the contrapositive and therefore follows at once.

Source : Link , Question Author : user79612 , Answer Author : Brian M. Scott

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