Find all pairs $(x, y)$ of integers that satisfy

$$
x^{2}+y^{2}+3^{3}=456 \sqrt{x-y} .
$$
Since the left side is an integer, the right side must also be an integer. The square root of an integer is either an integer or irrational (but never a non-integer fraction), so $\sqrt{x-y}$ must be an integer. The right side of the equation is divisible by 3, so the left side must also be. Thus, $3 \mid x^{2}+y^{2}$. But squares are always congruent to 0 or 1 modulo 3, so this can only happen if $x^{2} \equiv y^{2} \equiv 0 \bmod 3$. Therefore, $x$ and $y$ are both divisible by 3. Write $x=3 a$ and $y=3 b$ and substitute this in:

$$
9 a^{2}+9 b^{2}+3^{3}=456 \sqrt{3 a-3 b}
$$

Square roots of integers are generally either integers or irrational. But $\sqrt{3 a-3 b}$ cannot be irrational, because it is equal to $\frac{9 a^{2}+9 b^{2}+3^{3}}{456}$, so it must be an integer. Therefore, $3 a-3 b$ is a square, which is also divisible by 3, so it must be divisible by 9. We can now divide both sides of the equation by 9:

$$
a^{2}+b^{2}+3=152 \sqrt{\frac{a-b}{3}}
$$

Write $a-b=3 c^{2}$ and substitute $a=b+3 c^{2}$:

$$
9 c^{4}+6 c^{2} b+2 b^{2}+3=152 c
$$

Since all terms on the left side are positive, we must have $9 c^{4}+6 c^{2} b+2 b^{2}+3 \geq 152 c$. From $9 c^{4}+6 c^{2} b+2 b^{2}+3 \geq 152 c$ and $9 c^{4}+6 c^{2} b+2 b^{2}+3 \leq 152 c$, it follows that $c \leq 2$. Furthermore, $152 c$ is even, as is $6 c^{2} b+2 b^{2}$, so $9 c^{4}+3$ must also be even, which means $c$ must be odd. Therefore, the only possibility is $c=1$. Substituting this in, we get

$$
9+6 b+2 b^{2}+3=152
$$

or

$$
b^{2}+3 b-70=0
$$

This can also be written as $(b-7)(b+10)=0$. So $b=7$ or $b=-10$. In the first case, we get $a=b+3 c^{2}=10$ and thus $x=30$ and $y=21$. In the second case, we get $a=b+3 c^{2}=-7$ and thus $x=-21$ and $y=-30$. Checking shows that both pairs satisfy the equation. Therefore, the solutions are $(x, y)=(30,21)$ and $(x, y)=(-21,-30)$.