<s>[INST] If $n = 4k + 1$, does $4$ divide $n^2 -1$? How would I show that $4$ divides $n^2 -1$ if $n = 4k+1$? 
Is there more than one way to solve this?
 [/INST] For $k = 0$, $n = 1$, $n^2-1 = 0 = 4\cdot 0$.
Assume statement is true for some $k = h$. Let $(4h+1)^2-1 = 4a$.
For $k = h+1$, 
$$\begin{align*}
(4h+5)^2 - 1 &= (4 + 4h+1)^2 - 1\\
&= 4^2 + 2\cdot4(4h+1) +(4h+1)^2-1\\
&= 4^2 + 2\cdot4(4h+1) + 4a
\end{align*}$$
which is divisible by $4$.
Similarly for $k=h-1$.
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