I'm learning how to make rigorous proofs and I'm currently trying to prove this result:
Proposition: Let $n$ be an odd integer. Then there exists an integer $k$ such that $n^2=8k+1$.
I already prove the following result:
Propostion: Let $n$ be an integer. Then $n^2+n$ is an even integer.
So my proof is:
Let $n$ be an odd integer. Then there exists an integer $j$ such that $n=2j+1$. Then $n^2=(2j+1)^2=4j^2+4j+1=4(j^2+j)+1$ By the proposition stated, since $j$ is an integer, then $j^2+j$ is even. Then there exists an integer $k$ such that $j^2+j=2k$. Then $n^2=4(2k)+1=8k+1$. Hence there is an integer $k$ such that $n^2=8k+1$
I searched for different proofs and I found this one here: Proving that if $n$ is odd, then $n^2=8k+1 \text{ for some } k\in\mathbb{Z}$
I realized I could state that there exists an integer $k$ such that $2k=m(m+1)$. By substituting, we can conclude that $n^2=8k+1$. In terms of logic, what should be our approach to take this proof? Why can't we solve the equation to find that $2k=m(m+1)$? Thank you!