For what values of the parameter $t$ does the system of equations

$$
x+y+z+v=0, \quad(x y+y z+z v)+t(x z+x v+y v)=0
$$

have exactly one solution?
I. solution. Express $z$ from the first equation, substitute it into the second equation and arrange according to powers of $v$:

$$
v^{2}+(x+2 y-t y) v+\left(y^{2}+t x^{2}+t x y\right)=0
$$

The system of equations of the problem has a solution if and only if equation (3) has a solution. For any $t$, $(x, y, v)=(0,0,0)$ is a solution to (3), so (3) has exactly one solution if and only if the quadratic equation in $v$ has no solution for any $(x, y) \neq(0,0)$, i.e., its discriminant is negative:

$$
\begin{aligned}
D & =(x+2 y-t y)^{2}-4\left(y^{2}+t x^{2}+t x y\right)= \\
& =(1-4 t) x^{2}+2(2-3 t) x y+t(t-4) y^{2}0$ divided by (4) is equivalent

$$
(1-4 t) \frac{x^{2}}{y^{2}}+2(2-3 t) \frac{x}{y}+t(t-4)1$, then multiplying (2) by $\frac{1}{t}$ allows for similar reasoning.)

The left side of (2) can be transformed by squaring (1):

$$
(x y+y z+z v)+t(x z+x v+y v)=-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}+v^{2}+2(1-t)(x z+x v+y v)\right)
$$

Thus, (2) can be written in the form:

$$
A=x^{2}+y^{2}+z^{2}+v^{2}+2 s(x z+x v+y v)=0
$$

where $s=1-t$. We can write this sum as follows:

$$
A=(s x+z)^{2}+s(x+v)^{2}+(y+s v)^{2}+\left(1-s^{2}-s\right)\left(x^{2}+v^{2}\right)=0
$$

For $\frac{3-\sqrt{5}}{2}0$, so $x^{2}+v^{2}=0$, which means $x=v=0$. Then, since $s x+z=y+s v=0$, we also have $y=z=0$. Therefore, we have shown that if $\frac{3-\sqrt{5}}{2}<t<\frac{3+\sqrt{5}}{2}$, the system of equations of the problem has no non-trivial solution.