5. Determine the smallest natural number $N$, among the divisors of which are all numbers of the form $x+y$, where $x$ and $y$ are natural solutions to the equation $6 x y-y^{2}-5 x^{2}=7$. points)
Solution. Transform the equation by factoring the right-hand side

$6 x y-y^{2}-5 x^{2}-x^{2}+x^{2}=7 \Rightarrow 6 x(y-x)-(y+x)(y-x)=7 \Rightarrow(y-x)(6 x-y-x)=7 \Rightarrow$ $(y-x)(5 x-y)=7$.

Considering that the variables are natural numbers, and 7 is a prime number, we get

$$
\left\{\begin{array} { l } 
{ y - x = 7 , } \\
{ 5 x - y = 1 , \text { or } }
\end{array} \quad \left\{\begin{array} { l } 
{ y - x = - 7 , } \\
{ 5 x - y = - 1 , \text { or } }
\end{array} \quad \left\{\begin{array} { l } 
{ y - x = 1 , } \\
{ 5 x - y = 7 , \text { or } }
\end{array} \quad \left\{\begin{array}{l}
y-x=-1 \\
5 x-y=-7
\end{array}\right.\right.\right.\right.
$$

the second and fourth systems do not have natural solutions. The solution to the first system is the pair $(2 ; 9)$, and the third system is $(2 ; 3)$. Therefore, $x+y=11$ or $x+y=5$. The smallest natural number that has 5 and 11 as divisors is 55.

Answer: 55.