$2 \cdot 65$ When integers $x$ and $y$ are not both zero, find the minimum value of $f=|5 x^{2}+11 x y-5 y^{2}|$.
(Soviet Union University Mathematics Competition, 1977)
【Sol】When $x=1, y=0$
$$
f=\left|5 \cdot 1^{2}+11 \cdot 1 \cdot 0-5 \cdot 0^{2}\right|=5 \text {. }
$$

When $x, y$ are not both even, $f$ is odd.
When $x, y$ are both even, $\frac{x}{2}, \frac{y}{2}$ are integers, at this time we have
$$
f(x, y)>f\left(\frac{x}{2}, \frac{y}{2}\right) .
$$

If $\frac{x}{2}, \frac{y}{2}$ are still both even, the above process can continue. Therefore, the minimum value of $f$ is obtained when $x$, $y$ are not both even, at this time $f$ is odd, and the minimum value of $f$ can only be 1, 3, or 5.
If the minimum value of $f$ is 1, then there exist $x_{0}, y_{0}$ such that
$$
\begin{array}{l}
\left|5 x_{0}^{2}+11 x_{0} y_{0}-5 y_{0}^{2}\right|=1, \\
5 x_{0}^{2}+11 x_{0} y_{0}-5 y_{0}^{2}= \pm 1, \\
100 x_{0}^{2}+220 x_{0} y_{0}-100 y_{0}^{2}= \pm 20, \\
\left(10 x_{0}+11 y_{0}\right)^{2}-221 y_{0}^{2}= \pm 20 .
\end{array}
$$

Since $221=13 \cdot 17,20=13+7$, there exists an integer $t=10 x_{0}+11 y_{0}$, such that
$$
13 \mid t^{2} \pm 7
$$

Given
$$
\begin{array}{l}
t^{2}=(13 k)^{2} \equiv 0 \\
t^{2}=(13 k \pm 1)^{2} \equiv 1 \\
t^{2}=(13 k \pm 2)^{2} \equiv 4 \\
t^{2}=(13 k \pm 3)^{2} \equiv 9 \\
t^{2} \equiv(13 k \pm 4)^{2} \equiv 3 \\
t^{2} \equiv(13 k \pm 5)^{2} \equiv 12 \\
t^{2} \equiv(13 k \pm 6)^{2} \equiv 10
\end{array}
$$

At this time
$$
t^{2} \pm 7 \not \equiv 0
$$
$(\bmod 13)$, $(\bmod 13)$, $(\bmod 13)$, $(\bmod 13)$, $(\bmod 13)$, $(\bmod 13)$, $(\bmod 13)$, $(\bmod 13)$.
Therefore, the minimum value of $f$ cannot be 1.
If the minimum value of $f$ is 3, then there exist $x_{1}, y_{1}$ such that
$$
5 x_{1}^{2}+11 x_{1} y_{1}-5 y_{1}^{2}= \pm 3 \text {, }
$$

which means
$$
\left(10 x_{1}+11 y_{1}\right)^{2}-221 y_{1}^{2}= \pm 60 \text {. }
$$

Thus, there exists an integer $r$, such that $r=10 x_{1}+11 y_{1}$, and
$$
13 \mid t^{2} \pm 8 \text {. }
$$

From the above, this is also impossible. Therefore, the minimum value of $f$ is 5.