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[SOLVED] Factoring a^4+b^4

dwsmith

Well-known member
Feb 1, 2012
1,673
I am trying to write
\[
\frac{a^4+b^4}{a^2+b^2}
\]
with nothing higher than a power of two.

I know \(a^2+b^2 = (a + ib)(a - ib)\) and \(a^4 + b^4 = (a^2 + ib^2)(a^2 - ib^2)\), but I am to take the numerator down in farther in hopes of some cancelling in the denominator.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
It's a little-known fact that while $x^{2}+y^{2}$ does not factor over the reals, $x^{4}+y^{4}$ does. In fact,
$$x^{4}+y^{4}=(x^{2}+ \sqrt{2} xy+y^{2})(x^{2}- \sqrt{2} xy+y^{2}).$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
It's a little-known fact that while $x^{2}+y^{2}$ does not factor over the reals, $x^{4}+y^{4}$ does. In fact,
$$x^{4}+y^{4}=(x^{2}+ \sqrt{2} xy+y^{2})(x^{2}- \sqrt{2} xy+y^{2}).$$
So there won't be any cancelling. Since it ask for powers less than two, I could use my factoring just as well then?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
So there won't be any cancelling. Since it ask for powers less than two, I could use my factoring just as well then?
Sure. If you factor the numerator the way I have described, there won't be any powers written that are higher than $2$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Sure. If you factor the numerator the way I have described, there won't be any powers written that are higher than $2$.
I could also factor the numerator the way I factored it too. If not, why?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Perhaps use $(a^2+b^2)^2-2a^2b^2$?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
I could also factor the numerator the way I factored it too. If not, why?
Well, that would depend on whether you can factor over the complexes or not. If you can factor over the complexes, then you're fine. Otherwise, if you're going to factor, you'd have to use "my" factorization.