Factorize:

**great_math** · October 25th 2008, 11:10 PM

Can $a^4+b^4$ be factorized unlike $a^2+b^2$ ?

I think it can not be but there is a battle going on with me and my friend about this one.. So i just want to check it.

**11rdc11** · October 25th 2008, 11:29 PM

Originally Posted by great_math

Can $a^4+b^4$ be factorized unlike $a^2+b^2$ ?

I think it can not be but there is a battle going on with me and my friend about this one.. So i just want to check it.

How would you factor that? I think your friend is mixing up

$a^2-b^2 = (a+b)(a-b)$

**Chris L T521** · October 25th 2008, 11:32 PM

Originally Posted by great_math

Can $a^4+b^4$ be factorized unlike $a^2+b^2$ ?

I think it can not be but there is a battle going on with me and my friend about this one.. So i just want to check it.

If we were dealing with $a^3+b^3$ , $a^5+b^5$ and so on (the odd exponents), it can be factored.

Since we are dealing with $a^2+b^2$ or $a^4+b^4$ , it won't factor.

On the other hand, it is interesting to note that $a^6+b^6$ , $a^{10}+b^{10}$ and others can be factored, whereas $a^8+b^8$ , $a^{16}+b^{16}$ , $a^{32}+b^{32}$ , etc. cannot be factored....

In general, $a^k+b^k;~~k\geq 2$ can only be factored when $k\neq 2,~4,~8,~16,~32,...,~2^n;~~\forall~n\in\mathbb{N}$

--Chris

**mr fantastic** · October 25th 2008, 11:33 PM

Originally Posted by great_math

Can $a^4+b^4$ be factorized unlike $a^2+b^2$ ?

I think it can not be but there is a battle going on with me and my friend about this one.. So i just want to check it.

They can't be factorised using REAL factors. But they can be factorised using complex factors eg. a^2 + b^2 = (a + ib)(a - ib).

**great_math** · October 25th 2008, 11:59 PM

Can $a^4+b^4$ be factorized using complex numbers ?

**mr fantastic** · October 26th 2008, 12:13 AM

Originally Posted by great_math

Can $a^4+b^4$ be factorized using complex numbers ?

Yes:

$\left(a - \left[ \frac{\sqrt{2} + i \sqrt{2}}{2}\right] b\right) \, \left(a - \left[ \frac{\sqrt{2} - i \sqrt{2}}{2}\right] b\right)$ $\left(a <br /> + \left[ \frac{\sqrt{2} - i \sqrt{2}}{2}\right] b\right) \, \left(a + \left[ \frac{\sqrt{2} + i \sqrt{2}}{2}\right] b\right)$ .

**earboth** · October 26th 2008, 12:18 AM

Originally Posted by great_math

Can $a^4+b^4$ be factorized using complex numbers ?

$a^4+b^4=\left(a+\dfrac12 \sqrt{b} + \dfrac12 \cdot i \cdot \sqrt{b} \right)\left(a+\dfrac12 \sqrt{b} - \dfrac12 \cdot i \cdot \sqrt{b} \right)$ $\left(a-\dfrac12 \sqrt{b} + \dfrac12 \cdot i \cdot \sqrt{b} \right) \left(a-\dfrac12 \sqrt{b} - \dfrac12 \cdot i \cdot \sqrt{b} \right)$

EDIT: Too late ... again!

**NonCommAlg** · October 26th 2008, 12:26 AM

or on real numbers: $a^4 + b^4 = (a^2 + b^2 - \sqrt{2}ab)(a^2 +b^2 + \sqrt{2}ab).$ now try to factorize $a^8 + b^8$ over real numbers!

how many factors do you think $a^{2^n} + b^{2^n}$ will have over real numbers?

**mr fantastic** · October 26th 2008, 12:34 AM

Originally Posted by NonCommAlg

or on real numbers: $a^4 + b^4 = (a^2 + b^2 - \sqrt{2}ab)(a^2 +b^2 + \sqrt{2}ab).$ now try to factorize $a^8 + b^8$ over real numbers!

Yes, there's a nice re-arrangement that makes it fall out nicely.

For the OP: $a^4 + b^4 = (a^4 + b^4 + 2a^2 b^2) - 2a^2 b^2 = (a^2 + b^2)^2 - (\sqrt{2} \, a b)^2 = \, ....$

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