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?
 A: $$n^2-1=(n+1)(n-1)=(4k+2)4k=4p$$
with $p=k(4k+2)$.
A: if $n=4k+1$ then $n^2-1 = (4k+1)(4k+1)-1 = 16k^2 + 8k + 1 -1 = 16k^2 + 8k=4(4k^2 + 2k)$
Now clearly $4$ divides $4(4k^2 + 2k)$
Because $4k^2 + 2k$ is an integer , call it $m$
and so it is even more clear now that $4$ divides $4m$
and so $4$ does divide $n^2 -1$.
A: There are a couple of different ways of proving it, and I feel like even if I was the first to respond to you today, I would still be repeating what others have said (in response to questions similar to yours).
Although proof by contradiction is a valuable tool, it's kind of overkill here, in my opinion. The second most direct way is with simple algebraic rewriting. If $n = 4k + 1$, then $$n^2 = (4k + 1)^2 = (4k + 1)(4k + 1) = 16k^2 + 8k + 1.$$ It follows that $$\frac{n^2 - 1}{4} = \frac{16k^2 + 8k + \require{cancel}\color{red}{\cancel{1}} - \color{red}{\cancel{1}}}{4} = 4k^2 + 2k.$$ 
But the most direct way takes advantage of congruences, which I don't know if you've been taught yet. If $n = 4k + 1$, then $n \equiv 1 \pmod 4$. Therefore $n^2 \equiv 1 \pmod 4$ also, obviously leading to $n^2 - 1 \equiv 0 \pmod 4$.
A: 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 $k=h-1$.
A: $n = 4k + 1$, prove, $4 | n^2 - 1$.
If $k$ is an integer, $n$ will always be an odd number. 
Suppose $4$ does not divide $n^2 - 1$. 
$$n^2 \equiv a + 1 \pmod{4}$$
For some remainder $a \ne 0$
Because $n$ is odd, 
$$n^2 \equiv 1 \pmod{4}$$
Thus,
$$0 \equiv a \ne 0 \pmod{4}$$
Which is clearly false as $0 \equiv 0 \pmod{n}$ for any $n \ne 0$.
