## Task $2 / 76$

Determine all integer solutions $x ; y$ that satisfy the equation $3 x-2 y=1$!
Through equivalent transformation, the given equation follows:

$$
3 x=2 y+1 \quad ; \quad 3^{x}-3=2^{y}-2 \quad ; \quad 3\left(3^{x-1}-1\right)=2\left(2^{y-1}-1\right)
$$

This equation is certainly satisfied if \(3^{x-1}-1=0\) and \(2^{y-1}-1=0\). This leads to the first solution: \(x=1, y=1\). The verification confirms the correctness. If

$$
3^{x-1}-1 \neq 0 \quad \text { and } \quad 2^{y-1}-1 \neq 0
$$

then, due to the uniqueness of the prime factorization, it can only be

$$
3^{x-1}-1=2 \quad \text { and } \quad 2^{y-1}-1=3
$$

(The possibility of negative factors is excluded by the fact that the exponential function \(a^{x}\) is always greater than 1 for \(x \geq 0\), assuming \(a>1\).) Then it follows that \(3^{x-1}=3\) and \(2^{y-1}=4\), so \(x=2\) and \(y=3\).

The verification also confirms the correctness for this pair. Further solutions are excluded based on the solution process.