# 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.

• Can you prove that in an integral domain we have the factorization $$a^2-b^2=(a-b)(a+b)$$ Nov 20, 2014 at 16:36
• Do we start by rewriting $a^2=b^2$ as : $a^2+ab-ab-b^2=0$? Nov 20, 2014 at 16:52
• Is there a way to prove this without introducing adding and subtracting $ab$? Nov 20, 2014 at 17:04

Hint: the cancelation property is equivalent to $xy = 0 \implies x = 0$ or $y = 0$. And use the factorization $a^2-b^2 =(a-b)(a+b)$.

• Nice answer!!!! Nov 20, 2014 at 16:38
• So we start by writing $a^2-b^2=0$ as : $a^2+ab-ab-b^2=0$. But is there a way to show this without adding and subtracting $ab$? Nov 20, 2014 at 17:06
• Your problem is checking that $a^2-b^2 = (a-b)(a+b)$? You can think backwards, if you prefer: $$(a-b)(a+b) = a^2+ab-ba-b^2 = a^2-b^2,$$ since the ring is commutative. Nov 20, 2014 at 17:09
• Is there a way to show this without factoring though? I ask because I would've never thought about factoring. Nov 20, 2014 at 17:19
• I don't think so. I mean.. you could fix $y$ and define: $$f(x) = x^2-y^2 - (x+y)(x-y),$$ so $f(y) = 0$ and $$f'(x) = 2x - (x+y+x-y) = 0,$$ so $f \equiv 0$ and we get our result. But this is just to convince yourself in $\Bbb R$. Nov 20, 2014 at 17:24

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.