TITLE: Proving $(x^2 - y^2) = (x + y)(x-y)$
QUESTION [5 upvotes]: I am trying to prove from the field axioms that $(x^2 - y^2) = (x+y)(x-y)$. I am going to take for granted the definition of subtraction as the addition of a negation and that $x(-y) = (-x)y = -(xy)$. Here is my attempt.
We have:
\begin{align*}
(x+y)(x-y) & = x(x-y) + y(x-y) & & \text{distributive law} \\
& = x(x + (-y)) + y(x + (-y)) & & \text{definition of subtraction} \\
& = (xx + x(-y)) + (yx + y(-y)) & & \text{distributive law} \\
& = (x^2 + (-(xy))) + (yx + (-y^2)) & & \text{uniqueness of additive inverse; definition of square} \\
& = (x^2 + (-(yx))) + (yx + (-y^2)) & & \text{commutativity of multiplication} \\
& = (x^2 + \left(-(yx) + yx \right) + (-y^2) & & \text{associativity of addition} \\
& = (x^2 + 0) + (-y^2) & & \text{additive inverse axiom} \\
& = x^2 + (-y^2) & & \text{additive identity axiom} \\
& = x^2 - y^2 & & \text{definition of subtraction} 
\end{align*}
How does this look?

REPLY [4 votes]: Another way to approach it for the sake of curiosity:
\begin{align*}
x^{2} - y^{2} & = (x^{2} - xy) + (xy - y^{2})\\\\
& = x(x-y) + y(x - y)\\\\
& = (x+y)(x-y)
\end{align*}
Hopefully this helps!

REPLY [2 votes]: Looks good overall. I do have one nitpick:
At least for me, it is typically not given that the additive inverse is unique in fields, or that $x(-y) = -(xy)$ for sure. (Or the more general $(-1)(xy) = (-x)y = x(-y)$.)
I believe that technically needs to be shown if you want to go from the axioms.