If the square of a natural number is odd then this number is odd.

Question

My book says that

We represent $n$ as $n=2k+1$ where $k$ is from natural numbers or $k=0$

Then $n^2=(2k+1)^2=4k^2+4k+1$. We write $n^2=2(2k^2+2k)+1$ where $2k^2+2k$ is natural number or $0$ and we write it with $k_1$.

Now we can say that $n^2=2k_1+1$ where $k_1$ is a natural number or $0$. This proves what we wanted to know.

Now I don't get the last part of proof when we replaced the $2k^2+2k$. Can anyone help me?

Answer 1

The proof presented by your book actually proves the converse of the statement in the title:

If a number is odd, then the square of that number is odd.

As hinted in the comments, a quick way to prove the original statement in your title is to prove its contrapositive instead.

If a number is even, then the square of that number is even.

Answer 2

An other way :

Let $n=p_1\cdot ...\cdot p_m$ the prime decomposition of $n$. Since $$n^2=p_1^2\cdot ...\cdot p_m^2$$ is odd, $2$ don't divide any $p_i^2$, and thus $2$ don't divide any $p_i$ since $2\mid p_i\implies 2\mid p_i^2$. Therefore, $2$ don't divide $n$ and thus $n$ is odd.