## 6 Propagation Time for a Signal, Emitted and Intercepted by Satellites on a Space-Oriented Orbit
### Second Formulation of the Theorem for Four-Dimensional Uniformization
For the given cubic equation \(4\sigma^{3}-g_{2}\sigma-g_{3}=0\) with the first root (235), the discriminant is  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]6  
Let us assume that the roots of this polynomial are \(\alpha_{1},\beta_{1},\gamma_{1}\) and the roots of the polynomial \(4\sigma^{3}-\widehat{g}_{2}\sigma-\widehat{g}=0\) are \(\alpha_{2},\beta_{2},\gamma_{2}\). Then, the other two roots \(\beta_{1},\gamma_{1}\) can be found if the formulae for the s.c. \(n\)-th root of the number \(1\) is taken into account. This simple formulae is  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]7  
For \(n=3\), and by means of the formulae  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]8the roots \(\beta_{1},\gamma_{1}\) can be expressed as  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]9  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]0  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]1  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]2  
From the denominators of the two integrals in (234), we should have the equality of two cubic polynomials under a square root, the first expression for the polynomial obtained from the square of the second cubic polynomial, multiplied by \(\sqrt[3]{a}\)  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]3  
This expression can be written in the form  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]4  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]5  
The first root \(a^{\frac{2}{3}}\alpha_{2}\) of the cubic equation on the right-hand side of (242), analogously to the Cardano formulae (235) for the previous equation, can be expressed as  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]6  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]7  
From here, after multiplication of the right-hand side of (243) with \(a^{-\frac{2}{3}}\), only the root \(\alpha_{2}\) can be expressed as  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]8  
\[\overline{s}_{2}=a_{0}a_{4}-\frac{1}{4}a_{1}a_{3}+\frac{1}{12}a_{2}^{2}, \tag{223}\]9  
Now, it is important to note that \(\alpha_{2}\) is not another root of the cubic Equation (242)  
\[\overline{s}_{3}=\frac{1}{6}a_{0}a_{2}a_{4}+\frac{1}{48}a_{1}a_{2}a_{3}-\frac {a_{2}^{3}}{6^{3}}-\frac{1}{16}(a_{0}a_{3}+a_{1}^{2}a_{4}). \tag{224}\]0  
If one assumes that \(\alpha_{2}\) is another root (different from the roots \(\alpha_{1},\beta_{1},\gamma_{1}\)), then this would imply that the polynomial on the left-hand side of (242) should be of the fourth degree, which evidently will be a contradiction. Consequently, the root \(\alpha_{2}\) should coincide with one of the roots \(\alpha_{1},\beta_{1},\gamma_{1}\). Let us first assume that \(\alpha_{2}=\alpha_{1}\).