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.
 A: 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 $|N^*| \neq 5$.

A: This is the same as Dan's answer, just worked out.
$R$ has characteristic 2:

 Suppose $R$ has characteristic other than 2. Then for every unit, $u$, there is another unit $-u$, and if $u=-u$ then $2u=0$ a contradiction when $u$ is a unit. Hence the number of units of $R$ is infinite or even.

$R$ doesn't contain any big fields

 Suppose $R$ has characteristic 2. Then it may contain fields of characteristic 2. None of these fields can contain transcendentals, since there would be infinitely many units. None of these fields can be larger than $\mathbb{Z}/2\mathbb{Z}$ since such a field would contain a $2^n-1$st root of unity and $2^n-1$ only divides $5$ when $n=1$.

$R$ doesn't contain any real units:

 Consider $r \in R^\times$. Then $r^5=1$ so the subring generated by $r$ has 5 units and is isomorphic to $S=\mathbb{Z}/2\mathbb{Z}[x]/(f)$ where $f$ divides $x^5-1$. However, $x^5-1$ is a product of two distinct irreducibles, and hence $f$ is a product of distinct irreducibles, so that either (A) $f=x-1$ and $S = \mathbb{Z}/2\mathbb{Z}$, (B) $f=(x^5-1)/(x-1)$ and $S = \mathbb{F}_{16}$ with 15 units, a contradiction, or (C) $f=x^5-1$ and $S=\mathbb{Z}/2\mathbb{Z} \times \mathbb{F}_{16}$ with 15 units, a contradiction. Hence the only unit is $r=1$, contradicting the existence of 5 units.

