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!

"$n(n+1)$ is an integer multiple of 4"
We know we can write integers as prime factors, so $n(n+1)$ has 4 in its prime factors, and there are only two cases for this to happen:

*

*Both $n$ and $(n+1)$ are even, which is a refused case, because two successive numbers can't both be even.

*$n$ and $(n+1)$, one of them has to be odd while the other is a multiple of 4, and we know from the given that $n$ is not a multiple of 4, which leaves us with the case that $n$ is odd while $(n+1)$ is a multiple of 4.

For the second part of the proof, I'm going to represent $(n+1)$ as $4k$ such that $k \in \mathbb{Z}$, this means that $n=4k-1$, then we can do this:
$$n = 4k - 1 +4 - 4 \\ = (4k -4) + (4-1) \\ = 4(k-1) + 3$$
And according to closure property for addition on integers, $(k-1)$ is also an integer. Then we calculate the reminder:
$$\frac n 4 = \frac {4(k-1) +3} {4} = (k - 1) + \frac 3 4$$
Which means we are left with 3.
