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

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Discussion Overview

The discussion revolves around the simplification of the expression \( \frac{a^4+b^4}{a^2+b^2} \) with a focus on factoring \( a^4 + b^4 \) into components that do not exceed a power of two. Participants explore different factoring techniques and their implications for the expression.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests that \( a^2+b^2 \) can be factored as \( (a + ib)(a - ib) \) and \( a^4 + b^4 \) can be expressed as \( (a^2 + ib^2)(a^2 - ib^2) \), seeking further simplification.
  • Another participant presents a known factorization of \( x^4+y^4 \) as \( (x^{2}+ \sqrt{2} xy+y^{2})(x^{2}- \sqrt{2} xy+y^{2}) \), noting that this does not allow for cancellation in the expression.
  • There is a reiteration that the factorization provided does not lead to any terms higher than a power of two, which some participants agree upon.
  • One participant proposes an alternative approach using the identity \( (a^2+b^2)^2-2a^2b^2 \) as a potential method for factoring.
  • Another participant raises the consideration of whether factoring can be performed over the complex numbers, indicating that this affects the validity of different factoring approaches.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate methods for factoring \( a^4 + b^4 \) and whether cancellation is possible. There is no consensus on a single approach, and multiple competing views remain regarding the factorization techniques.

Contextual Notes

Participants note the limitations of factoring over the reals versus the complexes, which influences the applicability of certain factorizations. The discussion also highlights the need for clarity on the conditions under which different factorizations are valid.

Dustinsfl
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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.
 
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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,
$$x^{4}+y^{4}=(x^{2}+ \sqrt{2} xy+y^{2})(x^{2}- \sqrt{2} xy+y^{2}).$$
 
Ackbach said:
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?
 
dwsmith said:
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 $2$.
 
Ackbach said:
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?
 
Perhaps use $(a^2+b^2)^2-2a^2b^2$?
 
dwsmith said:
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.
 

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