Prove that square of number of form $2k+1$ is $8n+1$

Question

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 $8n+1$.

Answer 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 $8$.

Answer 2

$k$ and $k+1$ are consecutive integers, so either $k$ is even or $k+1$ is even. Hence $4k(k+1)$ is divisible by $8$.

Answer 3

Note that $(2k+1)^2 = 4k(k+1) +1 = 8\binom{k+1}{2}+1$.