Algebraic Proof

Prove algebraically that the square of any odd number is always 1 more than a multiple of 8.

Let n stand for any integer in your working.

Let k be a non-negative integer.

$\begin{array}{|rcll|} \hline n^2 &=& (2k-1)^2 \quad | \quad 2k-1 \text{ is any odd number } k > 0,\ k\in \mathbb{N} \\ &=& 4k^2-4k+1 \\ &=& 4k(k-1)+1 \quad | \quad \text{The product of two consecutive integers must be even so let k(k-1)=2m } \\ &=& 4\cdot 2m +1 \\ &=& 8m +1 \\ \mathbf{n^2} & \mathbf{=} & \mathbf{8m +1 \equiv 1 \pmod{8}} \\ \hline \end{array}$