I need guidance / correction for my proof. It's a little bit longer, but we really have to consider everything. If you find some issues / mistakes or have suggestions to improve it, please let me know! It helps alot!
Claim: All odd $$ n \in \mathbb{N} $$ can be displayed by the difference of two square numbers.
$$ k \in \mathbb{N0}$$ is a "square number" if there exists an $$ m \in \mathbb{N0}$$ that is $$m^2 = k $$
My proof (please critique if you think something is missing)Every $$n \in \mathbb{N}$$ can be displayed by $$(2k-1) = n^2$$
Proof by induction:
For n=1: $$ \sum_{n=1}^{1} = (2*1-1)=1 = 1^2 $$
Proof for n+1: $$ \sum_{n=1}^{n+1} = (2n-1) = (n+1)^2 $$
$$ \sum_{n=1}^{n+1} = (2n-1) + (2(n+1)-1) = (n+1)^2 $$ $$ \sum_{n=1}^{n+1} = (n^2) + (2(n+1)-1) = (n+1)^2 $$ $$ \sum_{n=1}^{n+1} = (n^2) + (2n+1) = (n+1)^2 $$
$$ \sum_{n=1}^{n+1} = (n^2 + 2n+1) = (n+1)^2 $$ End of induction proof.
From (n+1)^2 we can see the following:
$$ (n+1)^2 = (n^2+2n+1) $$$$ (n+1)^2 - n^2 = (2n+1) $$
Therefor, every $$ n \in \mathbb{N} $$ can be displayed by the difference of two square numbers with $$ (n+1)^2 - n^2 $$. Any critiques? The prof is really nitpicky and i would really appreciate some input to make the proof "bulletproof".
No induction is needed. Let $n$ be an odd positive integer; then there is a non-negative integer $k$ such that $n=2k+1$. And
$$n=2k+1=(k^2+2k+1)-k^2=(k+1)^2-k^2\,,$$
so $n$ is the difference of two squares — indeed, of two consecutive squares.