Polynomial equations system with an obvious solution—are there more?

Question

While doing some training in optimisation I was hit by the system of polynomial equations \begin{eqnarray} ta^6b+bd^2 & = & 2a^3\tag{1} \\[.67ex] tb^6c+ca^2 & = & 2b^3\tag{2} \\[.67ex] tc^6d+db^2 & = & 2c^3\tag{3} \\[.67ex] td^6a+ac^2 & = & 2d^3\tag{4} \\[.67ex] a^5 +b^5 +c^5 +d^5 & = & 4\tag{5} \end{eqnarray} where $\,a,b,c,d,t\,$ are real variables being strictly positive.

• The system is cyclic in $\,a,b,c,d$.
• The remaining variable $\,t\,$ gets easily isolated: Multiply $(1)$ through $(4)$ with the variables not present in the $t$-term, then sum up and use $(5)$ which yields $$t\cdot4abcd\:=\:a^3cd +b^3da +c^3ab +d^3bc\tag{6}\,.$$
• There's the solution with all variables $=1$.

Are there more solutions?
Otherwise, how could one prove that other solutions do not exist?

If further solutions exist, then not all $\,a,b,c,d\,$ are equal and WLOG one may assume $$a\geqslant b,c,d\quad\text{and}\quad 1<a<\sqrt[\uproot{2}5]{4}\approx 1.3195\,.$$ Furthermore one has $$abcd<1\quad\text{and}\quad t\cdot abcd<1\,,$$ both by application of 5-AM-GM to $\,abcd\cdot 1$, and termwise to the RHS of $(6)$, respectively, joint with $(5)$. Note that AM-GM is a strict inequality if not all arguments are equal.

Answer 1

YES, amazingly there is another real and positive solution to the system

$$\begin{aligned} ta^6b+bd^2 & = 2a^3\\ tb^6c+ca^2 & = 2b^3\\ tc^6d+db^2 & = 2c^3\\ td^6a+ac^2 & = 2d^3\\ a^5 +b^5 +c^5 +d^5 & = 4\end{aligned}$$

On the off-chance there was, I used Mathematica's Resultant function to reduce this to a single equation in one unknown. The extra solution is

$$a=(2x_1)^{1/5}=0.748744\dots\\ b=(2y_2)^{1/5}=1.24441\dots\\ c=(2x_2)^{1/5}=0.904281\dots\\ d=(2y_1)^{1/5}=0.706404\dots\\ t = \frac{2a^3-bd^2}{a^6b}=0.996756\dots\\ abcd =0.595186\dots<1\\ abcdt=0.593255\dots<1$$

where $x_1, x_2 \approx 0.117662, 0.302334$ are the two real roots of the $14$-deg,

$$\small{16 + 672 x + 12016 x^2 - 64176 x^3 - 678520 x^4 - 3705440 x^5 + 22284128 x^6 - 8541120 x^7 + 8010180 x^8 - 6259140 x^9 + 1893984 x^{10} - 212372 x^{11} + 34934 x^{12} - 6076 x^{13} + 343 x^{14}=0}$$

and $y_1, y_2 \approx 0.08795, 1.49205$ are the two real roots of the $14$-deg,

$$\small{16 - 1152 y + 24688 y^2 - 80400 y^3 - 796648 y^4 - 709040 y^5 + 950456 y^6 + 152880 y^7 + 2136 y^8 - 100380 y^9 + 15168 y^{10} - 19628 y^{11} + 3908 y^{12} + 1274 y^{13} + 343 y^{14} = 0}$$

The $a,b,c,d$ also satisfy the constraint,

$$\Big(\frac{a}{c}\Big)^3 = \frac{d}{b} $$

while $x_1,x_2$ satisfy the simple $7$th-deg,

$$ z^7 - 5z^6 + 7z^5 - z^4 - z^3 - z^2 - z - 1=0$$

as $\displaystyle z=\tfrac{1}{2}\Big(\frac{x_1}{x_2}+\frac{x_2}{x_1}\Big).$ P.S. I'm surprised at the simplicity of the $7$th-deg.