# Well-ordering proof

What is wrong with this "proof" of the following statement?

For every positive integer $$n,$$ the number $$n^2+n+1$$ is even.

Proof:

Let $$S$$ be the subset of positive integers $$n$$ for which $$n^2+n+1$$ is odd. Assume $$S$$ is nonempty.

Let $$m$$ be its smallest element.

Then $$m-1 \notin S,$$ so $$(m-1)^2+(m-1)+1$$ is even.

But $(m-1)^2+(m-1)+1 = m^2-m+1 = (m^2+m+1)-2m,$ so $$m^2+m+1$$ equals $$((m-1)^2+(m-1)+1)+2m,$$ which is a sum of two even numbers, which is even.

So $$m \notin S;$$ this is a contradiction. Therefore $$S$$ is empty, and the result follows.

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