$$
x^{4}+y^{4}=97(x+y)+Q, 0 \leq Qx\left(x^{3}-y^{3}\right)-(x+y) \\
\Rightarrow\left(98-y^{3}\right)(x+y)>0 \\
\Rightarrow 98>y^{3} \\
\Rightarrow 4 \geq y \ldots \ldots . . . .(1)
\end{array}
$$
$$
\begin{array}{l}
0 \leq Q \Rightarrow 0 \leq x^{4}+y^{4}-97(x+y) \\
\Rightarrow 97(5+y) \leq 625+y^{4} \\
\Rightarrow 97 y \leq 140+y^{4} \ldots . . . . . . . .(2)
\end{array}
$$

By trial and error, $y=1$ and $y=4$ satisfies (2)
So there are two possible pairs $(x, y)=(5,1)$ or $(5,4)$.
$5^{4}+1^{4}=626=(5+1) \times 104+2$, the quotient is not 97 .
$$
5^{4}+4^{4}=881=(5+4) \times 97+8, Q=8
$$
IS. 2 Given that $x$ and $y$ are positive integers. If $x^{4}+y^{4}$ is divided by $x+y$, the quotient is $P+13$ and the remainder is $Q$, find the value of $Q$.