Can $\displaystyle a^4+b^4$ be factorized unlike $\displaystyle a^2+b^2$ ?
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 $\displaystyle a^3+b^3$, $\displaystyle a^5+b^5$ and so on (the odd exponents), it can be factored.
Since we are dealing with $\displaystyle a^2+b^2$ or $\displaystyle a^4+b^4$, it won't factor.
On the other hand, it is interesting to note that $\displaystyle a^6+b^6$, $\displaystyle a^{10}+b^{10}$ and others can be factored, whereas $\displaystyle a^8+b^8$, $\displaystyle a^{16}+b^{16}$, $\displaystyle a^{32}+b^{32}$, etc. cannot be factored....
In general, $\displaystyle a^k+b^k;~~k\geq 2$ can only be factored when $\displaystyle k\neq 2,~4,~8,~16,~32,...,~2^n;~~\forall~n\in\mathbb{N}$
--Chris
Yes:
$\displaystyle \left(a - \left[ \frac{\sqrt{2} + i \sqrt{2}}{2}\right] b\right) \, \left(a - \left[ \frac{\sqrt{2} - i \sqrt{2}}{2}\right] b\right)$ $\displaystyle \left(a
+ \left[ \frac{\sqrt{2} - i \sqrt{2}}{2}\right] b\right) \, \left(a + \left[ \frac{\sqrt{2} + i \sqrt{2}}{2}\right] b\right)$.
$\displaystyle a^4+b^4=\left(a+\dfrac12 \sqrt{b} + \dfrac12 \cdot i \cdot \sqrt{b} \right)\left(a+\dfrac12 \sqrt{b} - \dfrac12 \cdot i \cdot \sqrt{b} \right)$ $\displaystyle \left(a-\dfrac12 \sqrt{b} + \dfrac12 \cdot i \cdot \sqrt{b} \right) \left(a-\dfrac12 \sqrt{b} - \dfrac12 \cdot i \cdot \sqrt{b} \right)$
EDIT: Too late ... again!
or on real numbers: $\displaystyle a^4 + b^4 = (a^2 + b^2 - \sqrt{2}ab)(a^2 +b^2 + \sqrt{2}ab).$ now try to factorize $\displaystyle a^8 + b^8$ over real numbers! how many factors do you think $\displaystyle a^{2^n} + b^{2^n}$ will have over real numbers?