Answer: 16511.

Solution. Such quadratic trinomials can be represented in the form $\left(x-5^{a}\right)\left(x-5^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a \geqslant b$. Expanding the brackets, we get $x^{2}-\left(5^{a}+5^{b}\right) x+5^{a+b}$. According to the condition

$$
\left\{\begin{array} { l } 
{ 5 ^ { a } + 5 ^ { b } \leqslant 1 2 5 ^ { 8 5 } , } \\
{ 5 ^ { a + b } \leqslant 1 2 5 ^ { 8 5 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
5^{a}+5^{b} \leqslant 5^{255} \\
a+b \leqslant 255
\end{array}\right.\right.
$$

Note that if the second inequality holds, then the first inequality is true except for one case $a=255, b=0$. For each value of $a$, we list the number of suitable values of $b$:

$$
\begin{aligned}
& a=255 \Rightarrow 0 \text { values of } b \text {; } \\
& a=254 \Rightarrow 2 \text { values of } b(b \in\{0 ; 1\}) \text {; } \\
& a=253 \Rightarrow 3 \text { values of } b(b \in\{0 ; 1 ; 2\}) \text {; } \\
& a=128 \Rightarrow 128 \text { values of } b(b \in\{0 ; 1 ; \ldots ; 127\}) \text {; } \\
& a=127 \Rightarrow 128 \text { values of } b(b \in\{0 ; 1 ; \ldots ; 127\}) \text {; } \\
& a=126 \Rightarrow 127 \text { values of } b(b \in\{0 ; 1 ; \ldots ; 126\}) \text {; } \\
& a=125 \Rightarrow 126 \text { values of } b(b \in\{0 ; 1 ; \ldots ; 125\}) \text {; }
\end{aligned}
$$

$$
\begin{gathered}
a=1 \Rightarrow 2 \text { values of } b(b \in\{0 ; 1\}) \\
a=0 \Rightarrow 1 \text { value of } b(b=0) .
\end{gathered}
$$

Summing up, we get $(2+3+4+\ldots 128)+(128+127+126+\ldots+1)=16511$ options.
3. Find the number of distinct reduced quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have at least one root, all their roots are powers of the number 5 with integer non-negative exponents, and their coefficients in absolute value do not exceed $122^{85}$.