finding the remainder of a variable n where n is not an integer multiple of 4 I started off by saying that $(n+1)$ must be a multiple of $4$ for $n(n+1)$ to be a multiple of 4. Hence $n$ when divided by $4$ would have a remainder of $3$
i.e $n = 4k+3$
$n(n+1)=(4k+3)(4k+3+1)
= 4(k^2+7k+3)$
Hence $n(n+1)$ is a multiple of 4
I'm not sure if this explanation is adequate and how to progress from here if it is not. Please advise, thanks!

We know that for a natural number $n$, $n$ and $n+1$ is odd-even pair. Which means that one of them is odd , the other is even.
So $n(n+1)$ is divisible by $4$ if and only if the even term is divisible by $4$. Now since $n$ is not the even term, $n+1$ must be. So, $n+1$=$4k$. Hence $n=4t+3$ where $t$ is any natural number.
