Factoring $x^{4} +1$, using real factoring to the second degree Factoring to the second degree using real numbers
$$x^{4} +1$$
I know that $ x^{4} +1=(x^{2} + i)(x^{2}-i).\;$ But these are complex but I thought using these in some kind of way? Got no where! 
And then I tried to guess, two solutions are $\pm (-1)^{1/4},\, $
which gave me $(x-\sqrt{i})(x+\sqrt{i}).\;$ But I have a feeling that I am really off here...
Hmm... Thinking! 
What kind of techniques do you use?  
 A: HINT :$$x^4+1+\color{red}{2x^2}-\color{red}{2x^2}=(x^4+2x^2+1)-2x^2=(x^2+1)^2-(\sqrt 2x)^2$$
A: Hint: Write $x^4 + 1 = (x^2 +1)^2 - (\sqrt{2}x)^2$
A: Your approach starts off perfectly fine: you do have
$$ x^4 + 1 = (x + \sqrt{\mathbf{i}})(x - \sqrt{\mathbf{i}})(x + \sqrt{-\mathbf{i}})(x - \sqrt{-\mathbf{i}}) $$
You just need to group them into conjugate pairs: (I assume you take the principal square roots)
$$ x^4 + 1 = \left( (x + \sqrt{\mathbf{i}})(x + \sqrt{\mathbf{-i}}) \right)\left( (x - \sqrt{\mathbf{i}})(x - \sqrt{\mathbf{-i}}) \right) $$
$$ x^4 + 1 = \left( x^2 + (\sqrt{\mathbf{i}} + \sqrt{-\mathbf{i}}) x + 1 \right) 
\left( x^2 - (\sqrt{\mathbf{i}} + \sqrt{-\mathbf{i}}) x + 1 \right) $$
The only simplification that remains is that you can compute
$$ \sqrt{\mathbf{i}} + \sqrt{-\mathbf{i}} = \sqrt{2}$$
A: Factor $x^4+1$ over $\mathbb{C}[x]$:
$$
\left(x-e^{\pi i/4}\right)\left(x-e^{3\pi i/4}\right)\left(x-e^{-3\pi i/4}\right)\left(x-e^{-\pi i/4}\right)
$$
Then pair the conjugate factors:
$$
\left(x-e^{\pi i/4}\right)\left(x-e^{-\pi i/4}\right)=x^2-\sqrt2\,x+1
$$
and
$$
\left(x-e^{3\pi i/4}\right)\left(x-e^{-3\pi i/4}\right)=x^2+\sqrt2\,x+1
$$
Therefore,
$$
x^4+1=\left(x^2-\sqrt2\,x+1\right)\left(x^2+\sqrt2\,x+1\right)
$$
A: Write the equation in the following form
$$(x^2+a_1x+a_2)(x^2+b_1x+b_2)=x^4+1$$
$$x^4+(a_1+b_1)x^3+(a_1b_1+a_2+b_2)x^2+(a_1b_2+a_2b_1)x+a_2b_2=x^4+1$$
By comparing the coefficients to the main equation we get
$a_1+b_1=0$
$a_1b_1+a_2+b_2=0$
$a_1b_2+a_2b_1=0$
$a_2b_2=1$
and solve for $a_1,a_2,b_1,b_2$
