TITLE: How to prove that from "Every infinite cardinal satisfies $a^2=a$" we can prove that $b+c=bc$ for any two infinite cardinals $b,c$?
QUESTION [13 upvotes]: Prove that if $a^2=a$ for each infinite cardinal $a$ then $b + c = bc$ for any two infinite cardinals $b,c$.

I tried $b+c=(b+c)^2=b^2+2bc+c^2=b+2bc+c$, but then I'm stuck there.

REPLY [9 votes]: Arturo's and Bill's answers assume that $b$ and $c$ are comparable, but that's not required. 
First I will show that if $b>1$ and $c>1$ we have $b+c\leq bc$. This is a fairly easy point, if you use the definitions of the operations. Take two disjoint sets $B$ and $C$ such that $|B|=b$, $|C|=c$; let $b_1,b_2\in B$ ($b_1\neq b_2$) and $c_1,c_2\in C$ ($c_1\neq c_2$). Define a function $F:B\cup C\to B\times C$ as follows:
$$F(p) = \begin{cases} (p,c_1), &p\in B\\
  (b_1,p), &p\in C\setminus\{c_1\}\\
  (b_2,c_2), &p=c_1\end{cases}$$
It's easy to verify  that this $F$ is an injection: Take any two elements $p,q\in B\cup C$ such that $p\neq q$. If one of them is $c_1$ observe that there is no other element that is sent to a pair that doesn't contain either $b_1$ or $c_1$, so their images are different. If $p,q\in B$ trivially $F(p)\neq F(q)$, (and likewise for two elements of $C$). If you have $p\in B$ and $q\in C\setminus\{c_1\}$, then $F(p)=(p,c_1)\neq(b_1,q)=F(q)$.
Now, you have shown in your question that $b+c=b+c+2bc\geq bc$. So we have that $b+c\leq bc$ and $bc\leq b+c$. The Cantor-Bernstein Theorem gives us that $b+c=bc$.