## Task A-3.2.

Determine all pairs of real numbers $(x, y)$ that satisfy the equation

$$
\left(4^{x}+1\right)\left(9^{y}+1\right)+70=10\left(2^{x}+1\right)\left(3^{y}+1\right)
$$
## Solution.

Let's introduce the substitution $a=2^{x}, b=3^{y}$.

The given equation becomes

$$
\begin{array}{cc}
\left(a^{2}+1\right)\left(b^{2}+1\right)+70=10(a+1)(b+1), & \\
a^{2} b^{2}+a^{2}+b^{2}+71=10(a b+a+b+1), & 1 \text { point } \\
a^{2} b^{2}+a^{2}+b^{2}-10 a b-10 a-10 b+61=0, & \\
\left(a^{2} b^{2}-12 a b+36\right)+\left(a^{2}+b^{2}+25+2 a b-10 a-10 b\right)=0, & 2 \text { points } \\
(a b-6)^{2}+(a+b-5)^{2}=0 . &
\end{array}
$$

Since the left-hand side is the sum of two non-negative real numbers, this sum is zero if and only if both addends are zero. Therefore,

$$
a b=6, \quad a+b=5
$$

The numbers $a$ and $b$ are the solutions to the quadratic equation $t^{2}-5 t+6=0$, whose solutions are 2 and 3.

In the first case, we have $a=2$ and $b=3$. Substituting into $a=2^{x}, b=3^{y}$ gives the solution $(x, y)=(1,1)$.

In the second case, we have $a=3$ and $b=2$. Substituting into $a=2^{x}, b=3^{y}$ gives the solution $(x, y)=\left(\log _{2} 3, \log _{3} 2\right)$.

Therefore, all solutions to the given equation are the pairs $(1,1)$ and $\left(\log _{2} 3, \log _{3} 2\right)$.

Note: A part of the solution that correctly identifies one of the ordered pairs earns 1 point per correct ordered pair. These points correspond to the last two points in the above scoring scheme.