MHB Factoring x^2 + y^2 | Sophie Germain's Identity

[Dethrone]

I came across an interesting website today, which stated that you can factor sums of squares IF 2AB is a perfect square known as the Sophie Germain’s Identity. Below is the website for your reference.

http://oakroadsystems.com/math/sumsqr.htm

I realize that if it is a perfect square, then the factorization would be nice and clean, but could you not do it also for non-perfect squares?

Example: factor $x^2 + y^2$

$A=x$
$B=y$
$2AB=2xy$

Therefore, $(x+y+\sqrt{2xy}) (x+y-\sqrt{2xy})$. If I were to expand this, I would get $x^2 + y^2$. Is this correct?

 

[Deveno]

I came across an interesting website today, which stated that you can factor sums of squares IF 2AB is a perfect square known as the Sophie Germain’s Identity. Below is the website for your reference.

http://oakroadsystems.com/math/sumsqr.htm

I realize that if it is a perfect square, then the factorization would be nice and clean, but could you not do it also for non-perfect squares?

Example: factor x^2 + y^2

A=x
B=y
2AB=2xy

Therefore, (x+y+sqrt(2xy)) (x+y-sqrt(2xy)). If I were to expand this, I would get x^2 + y^2. Is this correct?

Yes, that's true. The real question is: how useful is it?

For example, we can factor 25 as:

$(7 + \sqrt{24})(7 - \sqrt{24})$, but this factorization is not very easy to work with. This factors the rational number 25 into two irrational numbers, whereas the factorization:

$25 = 5\ast 5$

is much more practical.

So this raises the question: when is $\sqrt{2xy}$ rational?

 

[Dethrone]

You're right, it's not at all useful, just thought it was neat. I don't think sqrt(2xy) is rational, since sqrt(2) is an irrational number? Assuming that xy = 1.

 

[kaliprasad]

You're right, it's not at all useful, just thought it was neat. I don't think sqrt(2xy) is rational, since sqrt(2) is an irrational number? Assuming that xy = 1.

It is useful. One should realize that $x^2 + y^2$ is not the expression but the form

for exampe $a^4 + 4b^4$ is $x^2+y^2$ where $x=a^2$ and $y = 2b^2$ then $2xy= 4a^2b^2$ which is a perfect square

this can be utilized to factor

 

[Dethrone]

Ah, I see. This is unrelated to the original topic, but how may I go around factoring y^4 + y^2 + 1?

I know it is (y^2 + y + 1) (y^2 - y + 1), but not sure how to get there.

The only thing I notice is that there is some kind of difference of squares going around...
(y^2 + 1 + y) (y^2 + 1 - y)
= (y^2 + 1)^2 - y^2
= y^4 + 2y^2 + 1 - y^2

Is there an easier way to factor it, other than to rearrange the polynomial like what I just did?

 

[Prove It]

Ah, I see. This is unrelated to the original topic, but how may I go around factoring y^4 + y^2 + 1?

I know it is (y^2 + y + 1) (y^2 - y + 1), but not sure how to get there.

Let $\displaystyle \begin{align*} x = y^2 \end{align*}$ to get $\displaystyle \begin{align*} x^2 + x + 1 \end{align*}$ and then complete the square.

 

[Dethrone]

Ah, completing the square. Before I asked the question, I got to x^2 + x + 1, but I didn't know how to further factor it. Thanks!

EDIT: Wait... x^2 + x + 1 = (x + 1/2)^2 + 3/4...how do I further factor?

 

[Dethrone]

Still having a bit of trouble, actually.

If we let y^2 = x
then...
x^2 + x + 1, completing the square gives me (x+1/2)^2 + 3/4

How do I get (y^2 + y + 1) (y^2 - y + 1) from that?