2. Find all solutions to the equation

$$
m^{2}-2 m n-3 n^{2}=5
$$

where $m, n$ are integers.
Solution.

Transform the left side of the equation

$$
\begin{gathered}
m^{2}-2 m n+n^{2}-4 n^{2}=5 \\
(m-n)^{2}-(2 n)^{2}=5 \\
(m-n-2 n)(m-n+2 n)=5 \\
(m-3 n)(m+n)=5
\end{gathered}
$$

The integer 5 can be factored in four ways: $5=1 \cdot 5=5 \cdot 1=$ $(-1) \cdot(-5)=(-5) \cdot(-1)$. We get four systems of equations that need to be solved in integers. Solve by subtracting the first equation from the second.

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ m - 3 n = 1 , } \\
{ m + n = 5 , }
\end{array} \quad \left\{\begin{array} { l } 
{ 4 n = 4 , } \\
{ m + n = 5 , }
\end{array} \quad \left\{\begin{array}{l}
m=4 \\
n=1
\end{array}\right.\right.\right. \\
& \left\{\begin{array} { l } 
{ m - 3 n = 5 , } \\
{ m + n = 1 , }
\end{array} \quad \left\{\begin{array} { l } 
{ 4 n = - 4 , } \\
{ m + n = 1 , }
\end{array} \quad \left\{\begin{array}{l}
m=2, \\
n=-1
\end{array}\right.\right.\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ m - 3 n = - 1 , } \\
{ m + n = - 5 , }
\end{array} \quad \left\{\begin{array} { l } 
{ 4 n = - 4 , } \\
{ m + n = - 5 , }
\end{array} \quad \left\{\begin{array}{l}
m=-4 \\
n=-1
\end{array}\right.\right.\right. \\
& \left\{\begin{array} { l } 
{ m - 3 n = - 5 , } \\
{ m + n = - 1 , }
\end{array} \quad \left\{\begin{array} { l } 
{ 4 n = 4 , } \\
{ m + n = - 1 , }
\end{array} \quad \left\{\begin{array}{l}
m=-2, \\
n=1
\end{array}\right.\right.\right.
\end{aligned}
$$

Answer. $m=4, n=1 ; m=-4, n=-1 ; m=2, n=-1 ; m=-2, n=1$.