8. A three-digit number $n$ is a perfect square, and the sum of its digits is also a perfect square. There are $\qquad$ such three-digit numbers.

The sum of the digits of a three-digit number $n$ is a perfect square, and $n$ itself is a perfect square. How many such three-digit numbers are there?
【Answer】13
【Analysis】Comprehensive number theory: square numbers + estimation + remainder
For convenience, assume $a^{2}=n$. Since $31^{2}=961,32^{2}=1024$, we have $10 \leq a \leq 31$; the maximum sum of the digits of a three-digit number is 27, so the sum of the digits of $n$ could be $25, 16, 9, 4, 1$; the sum of the digits being $1: 10^{2}=100$, there is 1 such number;

The sum of the digits being 4: $11^{2}=121,20^{2}=400$, there are 2 such numbers;
The sum of the digits being 9: (square numbers that are multiples of 9)
$12^{2}=144,15^{2}=225,18^{2}=324,21^{2}=441,30^{2}=900$, there are 5 such numbers;
The sum of the digits being $16: (16 \equiv 7(\bmod 9)$, so $a \equiv 4(\bmod 9), a \equiv 5(\bmod 9))$ $13^{2}=169,14^{2}=196,22^{2}=484,23^{2}=529,31^{2}=961$, there are 5 such numbers; the sum of the digits being $25:$ (similarly $25 \equiv 7(\bmod 9)$, so $a \equiv 4(\bmod 9), a \equiv 5(\bmod 9))$ there are no such numbers; therefore, there are a total of $1+2+5+5=13$ numbers.