40. The squares of the 12th, 13th, and 15th terms of an arithmetic progression form a geometric progression. Find all possible common ratios of this progression.
40. Let's denote by $u_{12}, u_{13}, u_{15}$ the 12th, 13th, and 15th terms of an arithmetic progression, respectively. We know that the numbers $u_{12}^{2}, u_{13}^{2}, u_{15}^{2}$ form a geometric progression, i.e.,

$$
\frac{u_{13}^{2}}{u_{12}^{2}}=\frac{u_{15}^{2}}{u_{13}^{2}}
$$

Taking the square root of both sides of this equation, we get either

$$
\frac{u_{13}}{u_{12}}=\frac{u_{15}}{u_{18}}
$$

or

$$
\frac{u_{13}}{u_{12}}=-\frac{u_{15}}{u_{13}}
$$

Consider the first equation. For any arithmetic progression, $u_{13} = u_{12} + d, u_{15} = u_{13} + 2d = u_{12} + 3d$, where $d$ is the common difference of the progression. Therefore,

$$
\begin{aligned}
\frac{u_{12}+d}{u_{12}} & =\frac{u_{13}+2d}{u_{13}} \\
1+\frac{d}{u_{12}} & =1+\frac{2d}{u_{18}} \\
\frac{d}{u_{12}} & =\frac{2d}{u_{13}}
\end{aligned}
$$

Assume that $d \neq 0$, then

$$
\frac{u_{13}}{u_{12}}=2, \quad \frac{u_{13}^{2}}{u_{12}^{2}}=4
$$

i.e., in this case, the common ratio of the obtained geometric progression is 4.

Similarly, we can consider the case when the equation

$$
\frac{u_{13}}{u_{12}}=-\frac{u_{15}}{u_{13}}
$$

is satisfied. In this case, performing similar operations, we get the equations;

$$
\begin{gathered}
u_{12}^{2}+4 u_{12} d+d^{2}=0 \\
\left(\frac{d}{u_{12}}\right)^{2}+4 \frac{d}{u_{12}}+1=0 \\
\frac{d}{u_{12}}=-2 \pm \sqrt{3}
\end{gathered}
$$

Case a): $\frac{d}{u_{12}}=-2+\sqrt{3}$.

$$
\begin{gathered}
\frac{u_{13}}{u_{12}}=1+\frac{d}{u_{12}}=-1+\sqrt{3} \\
q=\left(\frac{u_{13}}{u_{12}}\right)^{2}=4-2 \sqrt{3}
\end{gathered}
$$

Case b): $\frac{d}{u_{12}}=-2-\sqrt{3}$.

$$
\begin{gathered}
\frac{u_{13}}{u_{12}}=1+\frac{d}{u_{13}}=-1-\sqrt{3} \\
q=\left(\frac{u_{13}}{u_{12}}\right)^{2}=4+2 \sqrt{3}
\end{gathered}
$$

Thus, from the equation

$$
\frac{u_{13}}{u_{12}}=-\frac{u_{15}}{u_{13}}
$$

it follows that $q$ can be either $4-2 \sqrt{3}$ or $4+2 \sqrt{3}$.

There is still the case $d=0$. Clearly, in this case, $q=1$.