Entry 31

This was inspired by Entry 30. My version is a solution to $$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2y)^{1/12}$$ can be given by \(x=4\arctan u_1\) and \(z=4\arctan u_2\) where the \(u_i\) are appropriate roots of the quartic$$a^2u^4+\tfrac{1}{a}(1-3a^2)u^3+3(1-a^2)u^2+a(3-a^2)u-1=0$$and \(a=\tan(y/4)\) for real \(0<y\leq1\). For example, let \(y=1\) so \(a=\tan(1/4)\approx 0.2553\), then \(u_1,u_2\) are the two real roots of the quartic.

Alternatively, let \(u_1,u_2\) be the same roots and define the Pythagorean triples$$\begin{aligned}p_1,\, p_2,\, p_3 &= 2u_1,\; u_1^2-1,\; u_1^2+1\\ q_1,\, q_2,\, q_3 &= 2u_2,\; u_2^2-1,\; u_2^2+1\\r_1,\, r_2,\, r_3 &= 2a,\; a^2-1,\; a^2+1\end{aligned}$$Then for some constant \(0<a<(-1+\sqrt{2})\), or \(1<a<(1+\sqrt{2})\), we have$$\left(\frac{p_1p_2(q_1^2-q_2^2)}{p_3^2q_3^2} \right)^{1/4}+\left(\frac{q_1q_2(p_1^2-p_2^2)}{p_3^2q_3^2} \right)^{1/4}=\left(\frac{4r_1r_2(r_1^2-r_2^2)}{r_3^4} \right)^{1/12}$$The equality holds, but I don't know why it works.