How to calculate the maximum voltage of square signal from Erms?
As you pointed out the equation for RMS is as follows:
\$V_{RMS} = \sqrt{V_{P}^2 + V_{DC}^2}\$
Where \$V_{P}\$ is voltage peak value of the AC component (half of \$V_{PP}\$), and \$V_{DC}\$ is the DC bias.
Because there are three variables there you can not simply determine the peak voltage of a DC square wave from its RMS alone. The amplitude of a wave with AC and DC components would be:
\$y = V_{P} + V_{DC}\$
So you would need to know one of these two values to determine the amplitude from the RMS. Lets assume you know the DC offset, then we can use our RMS equation and solve for \$V_P\$:
\$V_{P} = \pm \sqrt{V_{RMS}^2 - V_{DC}^2}\$
Next solve for \$V_P\$ in our displacement equation:
\$y = V_{P} + V_{DC}\$
\$V_{P} = y - V_{DC}\$
Now substitute this into the earlier equation and isolate:
\$V_{P} = \pm \sqrt{V_{RMS}^2 - V_{DC}^2}\$
\$y - V_{DC} = \pm \sqrt{V_{RMS}^2 - V_{DC}^2}\$
\$y = V_{DC} \pm \sqrt{V_{RMS}^2 - V_{DC}^2}\$
Thats as good as you get, that is the equation that will let you know the amplitude of the square wave from its RMS value and DC offset.