【Analysis】From the problem, we know that $c$ is at least $1111$, and $b$ is at most $99$. From $a^{2}+b=c$, we can see that $1111-99=1012$, so the minimum value of $a$ is 33, meaning $a$ could be one of the numbers 33, 44, 55, 66, 77, 88, 99. $c$ is a four-digit number with all digits the same, and it is no more than 100 greater than $a$'s square. We can solve the problem by trying each case.

【Solution】From the analysis, we have:
$a=33, a^{2}=1089, c=1111, b=22$, which meets the requirement;
$a=44, a^{2}=1936, c=2222, b=296$, which does not meet the requirement;
$a=55, a^{2}=3025, c=3333, b=308$, which does not meet the requirement;
$a=66, a^{2}=4356, c=4444, b=88$, which meets the requirement;
$a=77, a^{2}=5929, c=6666, b=737$, which does not meet the requirement;
$a=88, a^{2}=7744, c=7777, b=33$, which meets the requirement;
$a=99, a^{2}=9801, c=9999, b=198$, which does not meet the requirement;
There are three solutions that satisfy the requirements:
$$
(a, b, c)=(33,22,1111),(66,88,4444),(88,33,7777) .
$$
4. (10 points) Given that $a$ is a two-digit number with all digits the same, $b$ is a two-digit number with all digits the same, and $c$ is a four-digit number with all digits the same, and $a^{2}+b=c$. Find all $(a, b, c)$ that satisfy the conditions.