Problem 4. Find integer solutions of the equation: $2 x^{4}-4 y^{4}-7 x^{2} y^{2}-27 x^{2}+63 y^{2}+85=0$. Solution: Make the substitution: $x^{2}=a, \quad y^{2}=b$, then the equation becomes: $2 a^{2}-4 b^{2}-7 a b-27 a+63 b+85=0$

$2 a^{2}-(7 b+27) a-4 b^{2}+63 b+85=0$

$D=(9 b-7)^{2}$, the roots of the equation are $a_{1}=4 b+5$ and $a_{2}=\frac{17-b}{2}$

Case 1: $a=4 b+5 \Leftrightarrow x^{2}=4 y^{2}+5 \Leftrightarrow x^{2}-4 y^{2}=5 \Leftrightarrow(x-2 y) \cdot(x+2 y)=5$

There are four possible cases:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x - 2 y = 1 } \\
{ x + 2 y = 5 }
\end{array} \quad \left\{\begin{array} { l } 
{ x - 2 y = 5 } \\
{ x + 2 y = 1 }
\end{array} \quad \left\{\begin{array} { l } 
{ x - 2 y = - 1 } \\
{ x + 2 y = - 5 }
\end{array} \quad \left\{\begin{array}{l}
x-2 y=-5 \\
x+2 y=-1
\end{array}\right.\right.\right.\right. \\
& \left\{\begin{array} { l } 
{ x = 3 } \\
{ y = 1 }
\end{array} \quad \left\{\begin{array} { c } 
{ x = 3 } \\
{ y = - 1 }
\end{array} \quad \left\{\begin{array} { l } 
{ x = - 3 } \\
{ y = - 1 }
\end{array} \quad \left\{\begin{array}{c}
x=-3 \\
y=1
\end{array}\right.\right.\right.\right.
\end{aligned}
$$

Case 2: $a=\frac{17-b}{2} \Leftrightarrow 2 x^{2}=17-y^{2} \Rightarrow 2 x^{2} \leq 17 \Leftrightarrow x^{2} \leq 8.5 \Rightarrow$ $x^{2}$ can take the values 0, 1, or 4, calculate $y^{2}=17-2 x^{2} \Rightarrow$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7a16de7ae2a30fbf517g-3.jpg?height=316&width=665&top_left_y=1858&top_left_x=187)
Answer: $(3 ; \pm 1),(-3 ; \pm 1),(2 ; \pm 3),(-2 ; \pm 3)$