I was looking at the Agoh-Giuga Conjecture, namely, $$\sum_{i=1}^{n-1}i^{n-1}\equiv-1\pmod n\Leftrightarrow n\text{ is prime.}\tag1$$
I decided to see if I could prove it, expecting lots of difficulty; and I believe I was able to prove that if $n\equiv 3\pmod 4$ then Eq. $(1)$ $\equiv -1\pmod {2n}$.
Proof: Restating the conjecture, there exists an integer $k$ such that $$\begin{align}kn&=1+\sum_{i=1}^{n-1}i^{n-1} \\ &= 2 +\sum_{i=2}^{n-1}i^{n-1} \\ &= 2\big(1+2^{n-2}\big)+\sum_{i=3}^{n-1}i^{n-1}.\tag2\end{align}$$ Let's consider for $n>3$. Now any prime $>3$ is congruent to $\pm 1$ modulo $6$. Therefore, by order of substitution, $$k\big(6n\pm 1\big)=\text{Eq. }(2).$$ By rearranging, $$6nk-2\big(1+2^{n-2}\big)=\mp k+\sum_{i=3}^{n-1}i^{n-1}.$$ Note that if $n\equiv 3\pmod 4$ then the sum of strictly the powers is equal to $($the sum of an even number of odd numbers) $+$ $($the sum of an even number of even numbers). Therefore, that entire sum is even. Since the LHS is divisible by $2$, then that means $k$ also must be divisible by $2$.
Let $k=2j$, then we re-write that $$\begin{align}2jn&=1+\sum_{i=1}^{n-1}i^{n-1} \\ \therefore \sum_{i=1}^{n-1}i^{n-1}\equiv -1\pmod {2n}&\Leftrightarrow \text{$n$ is prime}\equiv 3\pmod 4.\end{align}$$ $$\quad\,\,\,\binom{\text{Assuming the conjecture}}{\text{is valid for all prime $n$}}$$ Of course $n>2$ since $2\not\equiv 3\pmod 4$. $\;\bigcirc$
Is my proof correct? Can it be written differently and in a better fashion? Is there anything new on this conjecture thus far?
Thank you in advance.