Q:

number of solutions of $x^2=x$ divides the number of invertible elements in a ring

Let $A$ be a commutative ring with odd number of elements. If $n$ is the number of solutions of the equation $x^2=x,x\in A$, and $m$ is the number of invertible elements of $A$, prove that $n$ divides $m$. Can somebody give me some tips, please? I do not know how to start the proof.Is  $2 \in U(A)$?

A:

Hint: $x^2=x$ iff $(2x-1)^2=1$.
Elaboration as requested in the comments. Consider the map $x\mapsto 2x-1$. Because $2$ is invertible in $A$, this map is a bijection from $A$ to itself. Therefore, by the hint above, it induces a bijection from the set of elements satisfying $x^2=x$ to the set of elements satisfying $y^2=1$. The latter, form a subgroup of the group of units and we're done by Lagrange.

