Number of units in a commutative ring In our Abstract Algebra class, we have just discussed the idea of a unit in a general ring and has posed the following question: "Prove that there cannot be a commutative ring with $1$ (the multiplicative identity) and $5$ units". 
My assumption was that there are no commutative rings that have an odd amount of units. This is because if $a$ is a unit, then there is a $b \in R$ such that $ab = ba = 1$. But, wouldn't this imply that $b$ would have to be a unit as well? And also, I know that if $a$ is a unit, then $-a$ is a unit as well. So, this would imply that units have to come in pairs and thus, we can never have an odd amount of units. However, this person has constructed a ring with an odd amount of units.
So, I guess my question is why can I not have $5$ units in a commutative ring? What about the commutativity gives me the fact that I cannot have $5$ units? 
 A: Your link actually gives a huge hint as to how to do this, as it shows that $a=-a$ in any counterexample, and more generally that $U(R)$ has no subgroups of order $2$.
Assume that $R$ is a communitive ring with unity and $5$ units. Our five units are $1, a, a^{-1}, b, b^{-1}$ for some $a,b$. To see that $a,a^{-1}$ are distinct, notice that, for all $x\in R$,
$$a=a^{-1}\Rightarrow xa=xa^{-1} \Rightarrow xa^{2}=x\Rightarrow a^2=1$$
Thus if $a\neq 1$, then $\{1,a\}\leq U(R)$ has order $2$, which is a contradiction. To see that $a\neq b, a\neq b^{-1}$ just note that if this wasn't to hold, then we just did a bad job of picking $b$, and there are two other units out there by assumption.
Now you need to show that no matter how multiplication is defined between the units, you can always derive a contradiction. Try looking at the possible values of $ab$ and then constructing an element whose square is $1$.
A: Assume there exists a commutative ring, $R$, with $|R^\times|=5$ (precisely 5 units).
Since $|R^\times|=5$ prime, this means  $R^\times$ a cyclic group generated by a single element, and we can view it as $\{ 1, u, u^2, u^3, u^4 \}$.
Now, $(-1)\cdot(-1)=1 \implies (-1)\in R^\times$ but only $1 \in R^\times$ satisfies $r^2=1$ (where $r\in R^\times$). In particular, $1=-1$ and we are in characteristic 2.
Using characteristic two, we can observe that, $(1+u+u^2)(u+u^2+u^4)=(u+u^6)+(u^2+u^2)+(u^4+u^4)+(u^3+u^3)+u^5 = 2(\ldots)+1 = 1$. And so $(1+u+u^2) \in R^\times$. We can't have $(1+u+u^2) \in \{ 1,u,u^2 \}$, since if $(1+u+u^2) = u^2$ for example, this would mean $1=-u \implies 1=u$ so $1$ and $u$ aren't distinct. So we must have $(1+u+u^2) \in \{u^3,u^4\}$.
By considering $r=(1+u+u^2+u^3+u^4)$. Specifically, $r^2 = 2(2r)+r = r$ (by expansion). Now, if $r$ is a unit $\implies r=1$ since $1$ is also only unit with property $r^2=r$. But, considering the two (virtually identical) cases,
Case 1: If $(1+u+u^2) = u^3 \implies r = u^3 + u^3 + u^4 \implies u^4 = r \implies u^4 =1$
Case 2: If $(1+u+u^2) = u^4 \implies r = u^3 + u^4 + u^4 \implies u^3 = r \implies u^3 =1$
Both cases leads to a contradiction and hence, there cannot exist a commutative ring with precisely 5 units!
