Modular quadratic equation question- Where did I go wrong? \begin{gather}
\frac{N^2+N}{2} \equiv 0 \pmod 4 \\
N^2+N\equiv 0\\
(N+\frac{1}{2})^2-\frac{1}{4}\equiv 0\\
4(N+\frac{1}{2})^2-1\equiv 0\\
2^2(N+\frac{1}{2})^2\equiv 1\\
(2N+1)^2\equiv 1 \\
2N+1 \equiv 1\\
2N=0\\
N=2,4,6,8...\\
2N+1 \equiv 3\\
2N=2\\
N=1,3,5,7...\\
\end{gather}
In other words I've shown $\forall N &gt;0 \to \frac{N^2+N}{2} \equiv 0 \pmod 4$, which is false e.g. for $N=3$,  $\frac{N^2+N}{2}$ has a remainder of 2
I'd like a correct solution as well. Thanks.
Your first error is that you you scaled by $\,2,\,$ which is not invertible $\!\bmod 4,\,$ so this will not yield an equivalent congruence. Rather, it yields necessary but not sufficient conditions on roots (so possibly  extraneous roots). See here for more on the insufficiency of unidirectional inferences.
To get an equivalent congruence we need to scale the modulus too, since $\,4\mid a/2\iff 8\mid a,\,$ so
$$(n^2+n)/2\equiv 0\!\!\!\pmod{4}\iff n^2+n\equiv 0\!\!\!\pmod{8}\qquad$$
Now we can complete the square as you did, but since this too involves scaling the modulus, this will end up being fruitless, leading back to where we started, namely
$$\begin{align} n^2+n&amp;\equiv 0\pmod{8}\\
\iff\ \ \ \ \ \ \ 4n^2+4n&amp;\equiv 0\pmod{32}\\
\iff 4n^2+4n+1&amp;\equiv 1\pmod{32}\\
\iff \ \ \ \ \ \, \color{#c00}{(2n+1)^2}&amp;\:\color{#c00}{\equiv 1}\pmod{32}\\
\iff (2n)(2n+2)&amp;\equiv 0\pmod{32}\\
\iff\ \ \ \ \ \ \  n(n+1)&amp;\equiv 0\pmod{8}
\end{align}\qquad$$
where we solved $\,\color{#c00}{a^2\equiv 1}\,$ by factoring the difference of squares $\,0\equiv a^2-1\equiv (a-1)(a+1).\,$  You can't simply take square roots as you did, e.g. $\,x^2\equiv 1\pmod{8}\,$ has $4$ roots $\,x\equiv 1,3,5,7$.
Instead, by $\,n,\,n\!+\!1\,$ coprime, $\,8\mid n(n\!+\!1)\iff 8\mid n\,$ or $\,8\mid n\!+\!1,\,$ thus we conclude that  $\,n(n+1)/2\equiv 0\pmod{4}\iff n\equiv 0,7\pmod{8}$
Finally, beware that modular fractions are well-defined (uniquely exist) only when they are writable with denominator coprime to the modulus, when $\,a/b := ab^{-1}.\,$ For more on modular fractions see here and here.
$(2N+1)^{2} \equiv 1  $
$2N+1 \equiv 1$
This step is wrong, for example: $(2*3+1)^2 \equiv 1$.