31. It is given that $x$ and $y$ are two real numbers such that
$$
(x+y)^{4}+(x-y)^{4}=4112 \text { and } x^{2}-y^{2}=16 \text {. }
$$

Find the value of $x^{2}+y^{2}$.
31. Answer: 34
By binomial theorem, one sees that
$$
(x+y)^{4}+(x-y)^{4}=2\left(x^{4}+6 x^{2} y^{2}+y^{4}\right)
$$

Thus the first equation becomes
$$
\begin{array}{l}
2\left(x^{4}+6 x^{2} y^{2}+y^{4}\right)=4112 \\
x^{4}+6 x^{2} y^{2}+y^{4}=2056 \\
\left(x^{2}-y^{2}\right)^{2}+8 x^{2} y^{2}=2056 \\
(16)^{2}+8 x^{2} y^{2}=2056 \\
x^{2} y^{2}=2256
\end{array}
$$

Therefore,
$$
\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}=1156
$$

Hence $x^{2}+y^{2}=\sqrt{1156}=34$.