How can we describe this factorization process?

Question

We know that, there exist infinite number of algebraic expressions, so that we can not factorise these with real coefficients. For example,

1. $a^2+b^2$

2. $a^2+b^2+c^2$

The simplest example, $a^2+b^2$ can not be factored over $\Bbb R$.

But, I want to consider the factorization like the one below.

$$a^2+b^2=(a+b)^2-2ab=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$$

But, what is the mathematical name of this type of factorisation? Is this a really factoring operation?

Edit:

Ah, I see that the, people misunderstood my question, maybe I was not clear.

I know that,

$$a^2+b^2=(a+b)^2-2ab=(a+b)^2-(\sqrt {2ab})^2=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$$

I am aware $$a^2-b^2=(a-b)(a+b)$$

But, my question is different here.

I wonder how this factoring operation is positioned in mathematics.

I mean, in which mathematical sense, this can be considered as factoring?

$$a^2+b^2=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$$

How can we describe/define this factorization process?

$$a^2+b^2=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$$ over $?$

Answer 1

Well, note that $a^2+b^2 = (a-bi)(a+bi)$ is also a factorization. Your factorization is a factorization over the ring $\mathbb{R}[\sqrt{a},\sqrt{b}]$ and over $\mathbb{R}[\sqrt{ab}]$. My factorization over the ring $\mathbb{Z}[i].$

The polynomial $x^2-3$ factors over $\mathbb{R}$ and over $\mathbb{Z}[\sqrt{3}]$ as $(x-\sqrt{3})(x+\sqrt{3}).$

Factorization depends on the ring one is factoring over. So, in that sense, the such factorizations are "positioned" in the context of the ring.

Answer 2

Let $u = a +b$ and let $v=\sqrt{2ab}$. Now we have $u^2-v^2=(u+v)(u-v)$ so it's just a quadratic in disguise.

Answer 3

Note that if $a=c^2$ and $b=2d^2$ then $a^2+b^2=(a+b)^2-2ab=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$ becomes $c^4+4d^4 =(c^2+2d^2)^2-4c^2d^2 =(c^2+2d^2-2cd)(c^2+2d^2+2cd) $ which is an unexpected factorization with integer coefficients.