1. Three non-zero real numbers form an arithmetic progression, and their squares taken in the same order form a geometric progression. Find all possible common ratios of the geometric progression.
Solution. Let the real numbers $a, b, c$ form an arithmetic progression, and the numbers $a^{2}, b^{2}, c^{2}$ form a geometric progression. Therefore, $2 b=c+a$ and $b^{4}=c^{2} a^{2}$. If we square the equation $2 b=c+a$, we get

$$
4 b^{2}=a^{2}+2 a c+c^{2}
$$

From the equation $b^{4}=c^{2} a^{2}$, we get $b^{2}=|a c|$, and if we substitute this into the previous equation, we have

$$
a^{2}+2 a c+c^{2}=4|a c|
$$

We will consider the two possible cases, i.e.,
a) $a$ and $c$ have the same sign
b) $a$ and $c$ have different signs

If $a$ and $c$ have the same sign, then the last equation takes the form $a^{2}-2 a c+c^{2}=0$, i.e., $a=c$. In this case, the ratio of the geometric progression is $q=1$.

If $a$ and $c$ have opposite signs, then the last equation takes the form $a^{2}+6 a c+c^{2}=0$ and since $a \neq 0$ it can be written in the form $\left(\frac{c}{a}\right)^{2}+6 \frac{c}{a}+1=0$. The solutions to the last quadratic equation in $\frac{c}{a}$ are $\frac{c}{a}=-3 \pm \sqrt{8}$. If the ratio of the geometric progression is $q$, then $\frac{c^{2}}{a^{2}}=q^{2}$, i.e., $q^{2}=(-3 \pm \sqrt{8})^{2}$. The numbers $a^{2}, b^{2}, c^{2}$ which form the geometric progression are positive, so according to this $q>0$. The positive solutions to the equation $q^{2}=(-3 \pm \sqrt{8})^{2}$ are $3-\sqrt{8}$ and $3+\sqrt{8}$.

Therefore, the required ratios are $1, 3-\sqrt{8}$, and $3+\sqrt{8}$.