Can this be considered factoring $a^2 + b^2$?

Question

We're told in middle school that there's no easy way to factor $a^2 + b^2$.

However, recently, and I believe this is common knowledge for more advanced mathematicians than me, I came with the following:

$$a^2 + b^2 = (a + b \mathit i)(a - b \mathit i)$$

So, I ask the better heads if this can be considered a factorization of $a^2 + b^2$.

Yes. It's a complex factorization, that's why it's not considered in middle school. — Jean-Claude Arbaut, Nov 13 '22 at 17:40
It's not admissible in $\mathbb{R}$, but it's surely a factorization in $\mathbb{C}$. — , Nov 13 '22 at 17:41
It is a valid factorisation in $\mathbb{C}$, but in general, especially at an elementary level, "factorisation" refers to finding factors in $\mathbb{Z}$ or at least $\mathbb{Q}$. — Deepak, Nov 13 '22 at 17:42
This is a correct factorization using complex number coefficients. Another factorization using complex numbers is discussed in the MSE question How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? — Dave L. Renfro, Nov 13 '22 at 17:42
This is considered a factorization in the $\mathbb{C}$ set, it is just too ahead of your grade level. If your teacher did say "there is no easy way" then she means that it isn't possible to do it in $\mathbb{R}$ set. — Kamal Saleh, Nov 13 '22 at 17:54

score 1 · Accepted Answer · answered Nov 13 '22 at 19:34

It can be considered a factorisation of $a^2 + b^2$, but it rarely would be without some context indicating that you're supposed to be working in the complex numbers.

I now would like to convince you that $a^2-b^2 = (a+b)(a-b)$ is a "better" factorisation than $a^2+b^2 = (a+bi)(a-bi)$.

For example, consider what happens when we work not in the integers (complex or not), but in the integers mod $n$; let's pick $n=5$ for concreteness. Then $$2 \equiv -3 = 1 - 4 \equiv 16 - 9 = \underbrace{4^2 - 3^2 = (4+3)(4-3)}_{\text{the factorisation!}} = 7 * 1 = 7 \equiv 2$$ still holds; $a^2 - b^2$ factorises just the same as it ever did. But what interpretation should we give to the symbol $i$ when we're working in the integers mod $5$? Well, I guess maybe it's the square root of $-1$, i.e. the square root of $4$, i.e. $i = 2$ (how weird); and indeed $a^2 + b^2 = (a + 2b)(a-2b)$ continues to hold.

What about if we're working mod $3$, where there is no square root of $-1$? In that case, we have to do what we do in the reals and adjoin a new object $i$ which we declare squares to $-1$, and it's not the case that $i$ is equal to an integer mod $3$.

It's all got messy, because now if we want to interpret what $(a+bi)(a-bi)$ means, we now have to work out whether there is a square root of $-1$ mod $n$.

So $a^2 - b^2$ factorises neatly in more general situations than $a^2+b^2$ does.

score -1 · Answer 2 · answered Nov 13 '22 at 19:20

When we say “factoring”, we usually mean taking a positive integer and writing it as the product of two or more positive integers that are not equal to 1. We have a^2-b^2 = (a+b)(a-b) which is a factoring when a and b are both integers. It is not a factoring if a = sqrt(1428) and b = sqrt(3000). The formula is correct, but a and b are not integers. Writing x as the product of two non-integers is trivial, let a ≠ 0 and b= x/a. That’s not a factoring.

What you found is a factoring in the Gauss numbers (complex numbers with real and complex part being integers) but we practically always want to factor in the integers.

Could you please use MathJax to format equations? – Gary Nov 23 '22 at 05:31 — Gary, Nov 23 '22 at 05:31

Can this be considered factoring $a^2 + b^2$?

2 Answers2