## Task 31/76

Determine all integer pairs $(x ; y)$ that satisfy the equation $16 x^{2}+2 x y+y^{2}=85$.


It is

$$
85=16 x^{2}+2 x y+y^{2}=15 x^{2}+(x+y)^{2} \geq 15 x^{2} \geq 0
$$

thus, $x^{2} \leq \frac{85}{15}<6, x^{2} \in\{0 ; 1 ; 4\}$.

Since the given equation yields integer values for $y$ when $x^{2}=4$, it follows that $x_{1}=2, x_{2}=-2$.

For $x^{2}=4$, we have $(x+y)^{2}=25$, i.e., $x+y= \pm 5$, and thus $y_{11}=3, y_{12}=-7, y_{21}=7, y_{22}=-3$.

Therefore, the equation has the four pairs $(2 ; 3),(2 ;-7),(-2 ; 7),(-2 ;-7)$ as solutions.