10. All positive integer solution pairs $(x, y)$ of the equation $x^{2}+y^{2}=208(x-y)$ are $\qquad$ .
10. $(48,32),(160,32)$.

Since 208 is a multiple of 4, the square of an even number is divisible by 4 with a remainder of 0, and the square of an odd number is divisible by 4 with a remainder of 1, so $x$ and $y$ must both be even.
Let $x=2a, y=2b$. Then
$$
a^{2}+b^{2}=104(a-b) \text {. }
$$

Similarly, $a$ and $b$ must both be even.
Let $a=2c, b=2d$. Then
$$
c^{2}+d^{2}=52(c-d) \text {. }
$$

Thus, $c$ and $d$ must both be even.
Let $c=2s, d=2t$. Then
$$
s^{2}+t^{2}=26(s-t) \text {. }
$$

Therefore, $(s-13)^{2}+(t+13)^{2}=2 \times 13^{2}$, where $s$ and $t$ are both even. So,
$$
\begin{array}{l}
(s-13)^{2}=2 \times 13^{2}-(t+13)^{2} \\
\leqslant 2 \times 13^{2}-15^{2}<11^{2} .
\end{array}
$$

Thus, $|s-13|$ could be $1, 3, 5, 7, 9$, and $(t+13)^{2}$ would be $337, 329, 313, 289, 257$, so $(t+13)^{2}$ can only be $289$. Therefore, $|s-13|=7$.
Thus, $(s, t)=(6,4),(20,4)$.
So, $(x, y)=(48,32),(160,32)$.