Prove in any integral domain, if $a^2=b^2$ then $a=\pm b$ Prove in any integral domain, if $a^2=b^2$ then $a=\pm b$
An integral domain is a commutative ring with unity having the cancellation property.  I don't see how I can use this in proving the statement.
 A: You have good answers. You are asking in the comments how one would arrive at such an answer. If I can take liberty in rephrasing: 

I never thought about factoring, would you have to do that -and- see how to factor $a^2-b^2$ in order to prove that this thing is true in every integral domain?

The main thing that "integral domain" gives you is that If $xy=0$ then $x=0$ or $y=0.$ To make use of this it would seem helpful to have "$=0$" statements so to change the question to 
Prove that in every integral domain If $a^2-b^2=0$ then $a-b=0$ or $a+b=0$. It seems the best chance is to use the defining property with $x=a-b$ and $y=a+b.$ Since you already have $x$ and $y$ you don't have to factor, you just have to multiply. 
Had the question been merely "When is $a^2=b^2$ in an Integral Domain?"  One would have to first get to $a^2-b^2=0$ and then factor (although knowing the solution $a=b$ does suggest the factor $a-b.$)

It is true that, for this line of reasoning, you have to be thinking of Integral Domain in this way... As far as the definitions, cancellation is the first property below and the rest are equivalent


*

*$xA=xB$ and $x \ne 0$ implies $A=B.$ 

*$x(A-B)=0$ and $x \ne 0$ implies $A-B=0.$

*$xy=0$ and $x \ne 0$ implies $y=0.$

*$xy=0$ implies $x=0$ or $y=0.$


(To a constructive mathematician the last is not equivalent... but that is another story.)
Sometimes the first form is more useful and sometimes, as in this case, the last form is more useful.
A: Hint: the cancelation property is equivalent to $xy = 0 \implies x = 0$ or $y = 0$. And use the factorization $a^2-b^2 =(a-b)(a+b)$.
