The Butterworth second-order low-pass filter has the following transfer function:
$$\frac{\omega_c^2}{s^2+\sqrt2s\omega_c+\omega_c^2}$$
We know that at \$s=\omega_c\$, the filter's attenuation is -3 dB, but mathematically, the only way to achieve that using the transfer function is to have a -1 coefficient in front of the \$\omega^2\$ term like so:
$$\frac{\omega_c^2}{\omega_c^2+\sqrt2\omega_c\omega_c-\omega_c^2}=\\\frac{\omega_c^2}{\omega_c^2(1+\sqrt2-1)}=\\\frac{1}{\sqrt2}=0.707$$
Thus, the gain at the cut-off frequency is \$20\cdot \log(0.707)=-3\ \mathrm{dB}\$
But every text book I researched shows the last term is +1, making the gain at the cut-off frequency equal to \$20\cdot\log(\frac{1}{(2+\sqrt2)})=20\cdot \log(0.2928)=-10\ \mathrm{dB}.\$
What went wrong?