Is n^2+n+3 Always Odd for Any Natural Number n?

[mathstudent88]

If n is a natural number then n²+n+3 is odd.

This is what I have and wanted to know if I was doing it right or not:

Let n be a member of the natural numbers. If n is even, then n=2k, k member of natural numbers, and n²+n+3
=(2k)²+2k+3
=4k²+2k+3
= 2(2k²+k+1)+1, where (2k²+k+1) is a member of the natural numers. This means that when n is even, n²+n+3 is odd.

If n is odd, then n=2j+1 where j is a member of the natural numbers
and n²+n+3
=(2j+1)²+(2j+1)+3
=4j²+4j+1+2j+4
=4j²+6j+5
=2(2j²+3j+2)+1, where (2j²+3j+2) is a memeber of the natural numbers. This means that when n is odd, n²+n+3.

Is this ok? Thanks for the help!

 

[Preno]

It looks ok, but it's overly complicated. A simpler argument is that n^2+n=n(n+1) is always even (because clearly one of n, n+1 is), therefore, n^2 + n + 3 is always odd.

 

[HallsofIvy]

I would say that what you give is a perfectly good proof and because it is yours it is the one you should submit. Of course, you should then be aware of Preno's simpler proof.