3. Two numbers x and y satisfy the equation $280 x^{2}-61 x y+3 y^{2}-13=0$ and are the fourth and ninth terms, respectively, of a decreasing arithmetic progression consisting of integers. Find the common difference of this progression.
Solution: We factorize the expression $280 x^{2}-61 x y+3 y^{2}$. For $y \neq 0$, we have $y^{2}\left(280\left(\frac{x}{y}\right)^{2}-61\left(\frac{x}{y}\right)+3\right)=280 y^{2}\left(\frac{x}{y}-\frac{3}{40}\right)\left(\frac{x}{y}-\frac{1}{7}\right)=(40 x-3 y)(7 x-y)$. This formula is also valid for all real numbers $y$. According to the problem, the integers $x$ and $y$ satisfy the equation $(40 x-3 y)(7 x-y)=13$. The integers $40 x-3 y$ and $7 x-y$ are divisors of the number 13, and the following cases are possible:

$$
\left\{\begin{array} { c } 
{ 4 0 x - 3 y = 1 3 , } \\
{ 7 x - y = 1 ; }
\end{array} \text { 2) } \left\{\begin{array} { c } 
{ 4 0 x - 3 y = - 1 3 , } \\
{ 7 x - y = - 1 ; }
\end{array} \text { 3) } \left\{\begin{array} { c } 
{ 4 0 x - 3 y = - 1 , } \\
{ 7 x - y = - 1 3 ; }
\end{array} \text { 4) } \left\{\begin{array}{c}
40 x-3 y=1, \\
7 x-y=13
\end{array}\right.\right.\right.\right.
$$

We solve each system using the method of addition. We multiply both sides of the second equation of the system by -3 and add it to the first:

$$
\left\{\begin{array} { l } 
{ 1 9 x = 1 0 , } \\
{ 7 x - y = 1 ; }
\end{array} \text { 2) } \left\{\begin{array} { l } 
{ 1 9 x = - 1 0 , } \\
{ 7 x - y = - 1 ; }
\end{array} \text { 3) } \left\{\begin{array} { c } 
{ 1 9 x = 3 8 , } \\
{ 7 x - y = - 1 3 ; }
\end{array} \text { 4) } \left\{\begin{array}{l}
19 x=-38 \\
7 x-y=13
\end{array}\right.\right.\right.\right.
$$

Systems 1) and 2) do not have integer solutions. Since the progression is decreasing, we have $x < y$, which is satisfied by the solutions of system 4): $x=-2$, $y=-27$. Thus, $a_{4}=-2, a_{9}=-27$, or $a_{1}+3 d=-2, \quad a_{1}+8 d=-27$. From this, we get $d=-5$.

Answer: $d=-5$.