@johncarlosbaez
you end up getting weird things like

a+ab+b=(a+b)^2+ab=
a+2ab+ba+b=a+3ab+2ba+b=
a+4ab+3ab+b=a+2ab+3ab+b (as 4=2) = ... =a+ba+b

Wow, that's cool! Check this out, @rogers.

@johncarlosbaez @alexthecamel excellent, that's even stronger than my last argument! Since this problem arose from an interest in commutativity, it's worth noting that even though \(ab \neq ba\) this shows that we have (for example) \((1+a)(1+b) = (1+b)(1+a)\).

@johncarlosbaez @alexthecamel
So, if \(a+b\) is a summand, we can not only reduce the coefficients of higher-order terms so that at most one is zero at each degree, we can actually reduce them to at most 1 (and they're all equivalent!):
\(a+ab + b = (a+ab)^2+b = a+2ab+aba+b\)
and by @alexthecamel 's argument, \(a+aba+b = a + (aba+b)^2 = a+aba + ba + ab + b = a + 2ab + aba + b.\)

@johncarlosbaez @alexthecamel A similar argument applies if we just have \(ab+ba\) as a summand, to conclude that \(ab+aba+ba = ab+bab+ba\). It's time I recounted with these results in mind!

@rogers
also true that
ab+aba+ba=(ab+ba)^2+aba
=ab+2aba+bab+ba=(ab+bab)^2+aba+ba
=2ab+2aba+2bab+ba
=2ab+aba+bab+ba
=2ab+ba

@alexthecamel I don't follow all of the steps in this one. Where does the second aba go in the second line?