Example 46 (2003 Bulgarian National Mathematical Olympiad) Given a sequence: $y_{1}=y_{2}=1$, $y_{n+2}=(4 k-5) y_{n+1}-y_{n}+4-2 k, n \geqslant 1$. Find all integers $k$ such that every term of the sequence is a perfect square.
Let $k$ satisfy the conditions of the problem.
From the given, we have
$y_{3}=2 k-2, y_{4}=8 k^{2}-20 k+13$.
Since $y_{3}$ is an even number, there exists an integer $a \geqslant 0$ such that $2 k-2=4 a^{2}$, then $k=2 a^{2}+1$, i.e., $k \geqslant 1$.
Thus, $y_{4}=32 a^{4}-8 a^{2}+1$.
When $a=0$, $k=1$.
When $a>0$, let $b (b \geqslant 0)$ be an integer satisfying $y_{4}=b^{2}$.
Thus, $16 a^{4}-8 a^{2}+1+16 a^{4}=b^{2}$, i.e., $\left(4 a^{2}-1\right)^{2}+\left(4 a^{2}\right)^{2}=b^{2}$.
Since $4 a^{2}-1$ and $4 a^{2}$ are coprime, $4 a^{2}-1, 4 a^{2}, b$ form a primitive Pythagorean triple. Therefore, there exist coprime positive integers $m, n$ such that
$$
\left\{\begin{array}{l}
4 a^{2}-1=n^{2}-m^{2}, \\
4 a^{2}=2 m n, \\
b=n^{2}+m^{2} .
\end{array}\right.
$$

From (1) and (2), we get
$n^{2}-m^{2}+1=2 m n$, i.e.,
$(n+m)^{2}-2 n^{2}=1$.
From (2), we have $n m=2 a^{2}$. Since $m, n$ are coprime, $m, n$ have different parities. But $m$ cannot be even, otherwise, it would imply $n^{2} \equiv-1(\bmod 4)$, which is impossible.
Therefore, $m$ is odd, and $n$ is even.
Thus, there exists an integer $t \geqslant 0$ such that $n=2 t^{2}$.
Using (4), we get
$$
2 n^{2}=8 t^{4}=(n+m-1)(n+m+1).
$$

So, $2 t^{4}=\frac{n+m-1}{2} \cdot \frac{n+m+1}{2}=u(u+1)$, where $u, u+1$ are consecutive integers. Since they are coprime, one must be a fourth power of some integer $\left(c^{4}\right)$, and the other must be twice a fourth power of some integer $\left(2 d^{4}\right)$. Thus,
$$
c^{4}-2 d^{4}= \pm 1 .
$$

If $c^{4}-2 d^{4}=1$, then
$$
d^{8}+2 d^{4}+1=d^{8}+c^{4} \text {, i.e., }\left(d^{4}+1\right)^{2}=\left(d^{2}\right)^{4}+c^{4} \text {. }
$$

The above equation is of the form $x^{4}+y^{4}=z^{2}$, which has been proven to have no integer solutions satisfying $x y z \neq 0$. Hence, $c=\pm 1, d=0$.
So, $u=0$, and $t=0$. Therefore, $n=a=0$. Contradiction.
If $c^{4}-2 d^{4}=-1$, then
$1-2 d^{4}+d^{8}=d^{8}-c^{4}$, i.e., $\left(d^{2}\right)^{4}-c^{4}=\left(d^{4}-1\right)^{2}$.
The above equation is of the form $x^{4}-y^{4}=z^{2}$, which also has no integer solutions satisfying $x y z \neq 0$. Thus, $d=\pm 1$, $c=\pm 1$.
For $u$, the only new value is $u=1$, so $t^{4}=1, n^{2}=4, n=2, m=1, a^{2}=1$, and $k=3$.
Thus, the only choices for $k$ are: $k=1$ and $k=3$.
When $k=1$, we get a periodic sequence: $1,1, 0,1,1, 0, \cdots$, which satisfies the constraints of the problem.
When $k=3$, the sequence is:
$y_{1}=y_{2}=1, y_{n+2}=7 y_{n+1}-y_{n}-2$.
Since $y_{3}=4=2^{2}, y_{4}=25=5^{2}, y_{5}=169=13^{2}$, we assume it is the square of the odd-indexed Fibonacci numbers.
If $\left\{u_{n}\right\}$ is the Fibonacci sequence, it is easy to see that $u_{n+2}=3 u_{n}-u_{n-2}$, and $u_{n+2} u_{n-2}-u_{n}^{2}=1$ for odd $n$.
Thus, $\left(u_{n+2}+u_{n-2}\right)^{2}=9 u_{n}^{2}$.
Therefore, $u_{n+2}^{2}=9 u_{n}^{2}-u_{n-2}^{2}-2 u_{n+2} u_{n-2}=7 u_{n}^{2}-u_{n-2}^{2}-2$.
This confirms our assumption.
Hence, $k=1$ and $k=3$.