3. (7 points) Find all pairs of natural numbers $x$ and $y$ such that

$$
\log _{2} a x+\log _{2} b y=\log _{2}\left(b x+a y+p_{1} p_{2}-1\right), \text { where } p_{1}, p_{2} \in \mathbb{P} \quad a, b>2
$$

In your answer, write the smallest possible value of $x+y$.
Solution: Let's get rid of the logarithms in the equation

$$
a x b y=b x+a y+p_{1} p_{2}-1
$$

Transform and factorize

$$
(a y-1)(b x-1)=p_{1} p_{2}
$$

Since $p_{1}, p_{2} \in \mathbb{P}$, we get four possible solutions

$$
\left\{\begin{array} { l } 
{ x = \frac { p _ { 1 } p _ { 2 } + 1 } { b } } \\
{ y = \frac { 2 } { a } }
\end{array} \quad \left\{\begin{array} { l } 
{ x = \frac { 2 } { b } } \\
{ y = \frac { p _ { 1 } p _ { 2 } + 1 } { a } }
\end{array} \left\{\begin{array} { l } 
{ x = \frac { p _ { 1 } + 1 } { b } } \\
{ y = \frac { p _ { 2 } + 1 } { a } }
\end{array} \quad \left\{\begin{array}{l}
x=\frac{p_{2}+1}{b} \\
y=\frac{p_{1}+1}{a}
\end{array}\right.\right.\right.\right.
$$

Since $a, b>2$ and $x, y \in \mathbb{N}$, the first two solutions are definitely not suitable. From the remaining two, we choose the one where $x, y$ are natural numbers and the sum $x+y$ is minimal.

Answers:

| Variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
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| Answer | 5 | 4 | 5 | 6 | 7 | 6 | 4 | 6 | 10 | 4 | 8 |