Butterworth filter approximation: derivation and output poles
Note that the Butterworth filter is only one of many possible approximations of an ideal low pass filter. It is a Taylor series approximation (at $\omega=0$) of the desired response, and, consequently, the magnitude response of the Butterworth filter is maximally flat at $\omega=0$.

Approximating the squared magnitude response (instead of the complex frequency response, or the magnitude response) has the advantage that the function to be approximated is an analytic real-valued function of a real-valued variable, which simplifies the approximation problem.

We have

$$|H(j\omega)|^2=H(j\omega)H^*(j\omega)=H(j\omega)H(-j\omega)=H(s)H(-s)\Big |_{s=j\omega}\tag{1}$$

because $H^*(j\omega)=H(-j\omega)$ for real-valued systems, i.e., for systems with a real-valued impulse response, or, equivalently, with real-valued transfer function coefficients.

Since for a Butterworth low pass filter 

$$|H(j\omega)|^2=\frac{1}{1+\omega^{2N}}\tag{2}$$

we have

$$H(s)H(-s)=\frac{1}{1+(s/j)^{2N}}=\frac{1}{1+(-s)^{2N}}\tag{3}$$

From the squared magnitude function we need to derive the transfer function $H(s)$, which is the function that we actually want to implement. From $(3)$ we know that the poles of $H(s)H(-s)$ (i.e., the zeros of the denominator of $(3)$) lie one a circle in the complex plane. Since we can only implement a causal and stable transfer function, we choose the $N$ poles in the left half-plane for $H(s)$, and consequently, the mirrored poles in the right half-plane are the poles of $H(-s)$.

Now we know $H(s)$ (i.e., its numerator and denominator coefficients), and, after having chosen a cut-off frequency and a filter structure, we can compute the values of the passive elements necessary to implement the given transfer function.