1. Find the maximum value of the expression $x+y$, where $x, y-$ are integer solutions of the equation $3 x^{2}+5 y^{2}=345$
# Solution

Notice that 345 and $5 y^{2}$ are divisible by 5, so $3 x^{2}$ must also be divisible by 5. Therefore, $\quad x=5 t, t \in Z$. Similarly, $y=3 n, n \in Z$. After simplification, the equation becomes $5 t^{2}+3 n^{2}=23$. Therefore, $t^{2} \leq \frac{23}{5}$, $n^{2} \leq \frac{23}{3}$ or $|t| \leq 2,|n| \leq 2$. By trying the corresponding values of $t, n$, we get that $|t|=2,|n|=1$ or

$$
\left\{\begin{array}{c}
x_{1}=-10 \\
y_{1}=-3
\end{array},\left\{\begin{array}{c}
x_{2}=-10 \\
y_{2}=3
\end{array},\left\{\begin{array}{l}
x_{3}=10 \\
y_{3}=-3
\end{array},\left\{\begin{array}{c}
x_{4}=10 \\
y_{4}=3
\end{array}\right.\right.\right.\right.
$$

The maximum value of the expression $x+y$ is $10+3=13$.

Answer. 13.