If n is an integer prove (n+3)^(2)-n^(2) is never even.

Let us begin by simplifying the expression:

(n+3)2 - n2  = (n+3)(n+3) - n2

= n2 + 6n + 9 - n2 (expanded brackets)

= 6n + 9 (collected like terms)

= 3(2n+3) (taken out a factor of 3)

Now we can consider this simpler equivalent expression.

3 is an odd number

2n is even thus 2n+3 is odd (even plus odd is always odd)

so we have an odd*odd which is always odd, thus never even and we are done.

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