Problem 1347

As $x$ and $y$ occur under square root, only non-negative solutions can exist. Bringing $\sqrt{x}$ to the right, squaring both sides and collecting similar terms gives $2019 = y + 2\sqrt{xy}$ which is equivalent to the initial equation. Thus $2\sqrt{xy}$ is an integer. If $x = 0$ then $y = 2019$, the case $y = 0$ leads to contradiction. Assume in the rest that both $x$ and $y$ are positive.
Let $a^2$ and $b^2$ be the largest perfect squares dividing $x$ and $y$, respectively; then $x = a^2c$ and $y = b^2c'$ where both $c$ and $c'$ are square-free. As $a^2c \cdot b^2c'$ is a perfect square, also $c \cdot c'$ must be a perfect square; this is possible only if $c$ and $c'$ have the same prime factors, i.e., $c = c'$. Hence
$$2019 = b^2c + 2\sqrt{a^2b^2c^2} = b^2c + 2abc = bc(2a + b).$$
Since $2019 = 3 \cdot 673$ where both factors are prime, we have the following cases, taking into account that $b < 2a + b$:
* $b=1, c=1, 2a+b=2019$, implying $a=1009, x=1009^2$ and $y=1$;
* $b=1, c=3, 2a+b=673$, implying $a=336, x=336^2 \cdot 3$ and $y=3$;
* $b=1, c=673, 2a+b=3$, implying $a=1, x=673$ and $y=673$;
* $b=3, c=1, 2a+b=673$, implying $a=335, x=335^2$ and $y=9$.

As $x$ and $y$ occur under square root, only non-negative solutions can exist. Bringing $\sqrt{x}$ to the right, squaring both sides and collecting similar terms gives $2019 - y = 2\sqrt{xy}$ which is equivalent to the initial equation. Hence $y \le 2019$. After squaring once more and rearranging terms, we obtain $2019^2 = y(2 \cdot 2019 - y + 4x)$. Thus $y \mid 2019^2$. Case study leads to the following solutions: (1) $y = 1, x = 1009^2$; (2) $y = 3, x = 336^2 \cdot 3$; (3) $y = 9, x = 335^2$; (4) $y = 673, x = 673$; (5) $y = 2019, x = 0$.