Prove that square of number of form $2k+1$ is $8n+1$ Need show that $\forall k, \exists n \in \mathbb{Z}, (2k+1)^2 =8n+1$.
 $4k^2 + 4k +1 = 4k(k+1) +1$.
Although, $2k+1$ is an odd number, but $k$ can be any integer. So, $k+1$ can be an odd or an even number. Hence, not sure of getting the number of the form $8n+1$.

Recursively By induction : if $n$ is odd then $$\begin{align}n^2  &= (n-2 +2)^2 \\ &= (n-2)^2+ 4(n-2) + 4 \\ &= (n-2) ^2 + 4(n-1)\end{align}$$ And since $n-1$ is even, $4(n-1)$ is a multiple of $8$.
