MHB Factor a^4+b^4: Simplify \(a^4 + b^4 \)

[Dustinsfl]

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]

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,
$$x^{4}+y^{4}=(x^{2}+ \sqrt{2} xy+y^{2})(x^{2}- \sqrt{2} xy+y^{2}).$$

 

[Dustinsfl]

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]

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 $2$.

 

[Dustinsfl]

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?

 

[I like Serena]

Perhaps use $(a^2+b^2)^2-2a^2b^2$?

 

[Ackbach]

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.