Group of invertible elements of a ring has never order $5$ 
Let $R$ be a ring with unity. How can I prove that group of invertible elements of $R$ is never of order $5$?

My teacher told me and my colleagues that problem is very hard to solve. I would be glad if someone can provide me even a small hint because, at this point, I have no clue how to attack the problem.

Here are a couple of ideas:


*

*$-1 \in R$ is always invertible. If $-1 \neq 1$, then it follows that $R^*$ should have even order, a contradiction. Therefore, $1=-1$ in ring $R$, so in fact $R$ contains a subfield isomorphic to $\mathbb{F}_2$.

*Let $a$ be the generator of $R^*$. Consider the subring $N \subseteq R$ generated by $1$ and $a$. Then in fact $N \simeq \mathbb{F}_2[x] / (f(x))$, where $\mathbb{F}_2[x]$ is the polynomial ring over $\mathbb{F}_2$, and $f(x)$ is some polynomial from that ring.

*Since $a^5=1$, it follows that $f(x)$ divides $x^5+1$. This leaves only finitely many options for $f$, and therefore for $N$. Then you can deal with each case separately and see that in each case $|N^*| \neq 5$.

