$8 \cdot 57$ a (finite) sequence of the same non-zero digit can be the ending of a perfect square. Find the maximum length of such a sequence, and the smallest square number whose ending is such a sequence.
(31st Putnam Mathematical Competition, 1970)
[Solution] Let $x$ be an integer, then the unit digit of $x^{2}$ can only be $0,1,4,9,6,5$. The case where the unit digit of $x^{2}$ is 0 has been excluded by the assumption of the problem.
If the last two digits of $x^{2}$ are the same, then they can only be $11,44,55,66,99$.
When the last two digits of $x^{2}$ are $11,55,99$,
$$
x^{2} \equiv 3(\bmod 4),
$$

which is impossible.
When the last two digits of $x^{2}$ are 66,
$$
x^{2} \equiv 2(\bmod 4),
$$

which is also impossible.
Next, we discuss the case where the last few digits of the square number are all 4.
If the last four digits of $x^{2}$ are 4, i.e.,
$$
x^{2} \equiv 4444(\bmod 10000),
$$

then
$$
x^{2} \equiv 12 \quad(\bmod 16) \text {. }
$$

In this case, from
$$
x^{2}=16 k+12=4(4 k+3)
$$

we know that it is also impossible.
Therefore, the last digits of $x^{2}$ can have at most three 4s.
Notice that
$$
38^{2}=1444 .
$$

Thus, the longest such sequence is 3, i.e., the last three digits are 4.
And 1444 is the smallest square number that meets the requirement.