Solve the equation

$$
8 x^{3}+8 x^{2} y+8 x y^{2}+8 y^{3}=15\left(x^{2}+y^{2}+x y+1\right)
$$

in the set of integers.
We transform the equation to the following one

$$
\left(x^{2}+y^{2}\right)(8 x+8 y-15)=15(x y+1)
$$

Since the right side is divisible by 3, then $3 \mid \left(x^{2}+y^{2}\right)(8 x+8 y-15)$. But if $3 \mid \left(x^{2}+y^{2}\right)$, then $3 \mid x$ and $3 \mid y$, which will give $15(x y+1)$ and $3 \mid (x y+1)$, which is impossible. Hence $3 \mid (x+y)$ and 3 does not divide $x$ or $y$. Without loss of generality we can assume that $x=3 a+1$ and $y=3 b+2$. Substituting in the equation, we obtain

$$
\left(x^{2}+y^{2}\right)(8(a+b)+3)=5(x y+1)
$$

Since $x y+1 \equiv 0 \pmod{3}$, we conclude that $3 \mid (a+b)$.

Now we distinguish the following cases:

- If $a+b=0$, then $x=3 a+1$ and $y=-3 a+2$ from which we get

$$
\left(9 a^{2}+6 a+1+9 a^{2}-12 a+4\right) \cdot 3=5\left(-9 a^{2}+3 a+3\right) \text { or } 3 a^{2}-a=0
$$

But $a=\frac{1}{3}$ is not an integer, so $a=0$ and $x=1, y=2$. Thus, by symmetry, we have two solutions $(x, y)=(1,2)$ and $(x, y)=(2,1)$.

- If $a+b \neq 0$, then $|8(a+b)+3| \geq 21$. So we obtain

$$
\left|\left(x^{2}+y^{2}\right)(8(a+b)+3)\right| \geq 21 x^{2}+21 y^{2} \geq|5 x y+5|
$$

which means that the equation has no other solutions.

## Geometry

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