Everything Can Be Cancelled Off Too?

It can be written into $\large i^{\frac{n(n+1)}{2}} = -1$ Which is only satisfy when the exponents divided by 4, leaves the reminder of 2. This may be written into $\frac{n(n+1)}{2} \equiv 2 \mod{4}\Rightarrow 8|(n^2+n-4)$ The positive integer $n$ satisfying this congruence is in the form of $n = (8j+\frac{7}{2})\pm\frac{1}{2};\:j\in\mathbb{N}$ In the above question, $12$ is an answer, as it may be written in the form of $\displaystyle n = (8(1)+\frac{7}{2})+\frac{1}{2}$

$\Large\ i^{\frac{n(n+1)}{2}}=-1.\ \ \ \therefore\ \dfrac{n(n+1)}{2}\ \equiv\ 2\ (mod\ 4).\\ That\ is\ \dfrac{n(n+1)}{2}\ divided\ by\ 4\ should\ give\ remainder\ as\ 2.\\ \implies\ n(n+1)\ \ divided\ by\ 8\ should\ give\ remainder\ as\ 4.\\ \dfrac{12*13} 8=19\frac 1 2. \ remainder\ is\ \frac 1 2*8=4.\\ \text{So 12 is the answer. Next number would be 12+8=20. So 13,14,15 can not be the answer.}$

We have $i^{\frac{n(n+1)}{2}}=-1$ if $n$ or $n+1$ is divisible by 4 but not by 8. This is the case for $n=\boxed{12}$ only.