Solution 91eaaf49-2179-4a17-8321-abb4b37a629e

Points: 1 pt

Prime factorizing,
\[
1183=7\cdot169=7\cdot13^2.
\]
Substitute
\[
x=7i^2,\qquad y=7j^2,
\]
where $i,j$ are positive integers. The original equation becomes
\[
i\sqrt7+j\sqrt7=13\sqrt7
\quad\Longrightarrow\quad i+j=13.
\]
Now,
\[
i^2+j^2=(i+j)^2-2ij=169-2ij.
\]
To minimize this expression, we maximize $ij$. Since $i+j=13$, the
product is largest when $i$ and $j$ are as close as possible:
$i=6$ and $j=7$. Hence the minimum value is
\[
x+y=7(i^2+j^2)=7(169-2\cdot6\cdot7)=595.
\]

Final answer:
595