850. Solve the equation in integers

$$
\sqrt{x}+\sqrt{y}=\sqrt{98}
$$
$\triangle$ From the equation, it is clear that

$$
0 \leqslant x \leqslant 98, \quad 0 \leqslant y \leqslant 98
$$

Let's represent it in the form $\sqrt{y}=\sqrt{98}-\sqrt{x}$ and square the last equation:

$$
y=98+x-2 \sqrt{98 x}, \quad y=98+x-14 \sqrt{2 x}
$$

## From this

$$
2 x=4 a^{2}, \quad x=2 a^{2},
$$

where $a$ is a non-negative integer. Since $x \leqslant 98$, then

$$
2 a^{2} \leqslant 98, \quad a^{2} \leqslant 49, \quad 0 \leqslant a \leqslant 7 .
$$

Therefore,

$$
a=0,1,2,3,4,5,6,7
$$

and accordingly

$$
x=0,2,8,18,32,50,72,98
$$

For each of these values of $x$, we find the corresponding value of $y$. Answer: $(0 ; 98),(2 ; 72),(8 ; 50),(18 ; 32),(32 ; 18),(50 ; 8),(72 ; 2),(98 ; 0)$.