A five-digit number whose square has its last five digits equal to the number.

Instead of a five-digit number, let's find a two-digit number. We want $x^2 - x = x(x-1)$ to be a multiple of 100. So it's a multiple of 4 and of 25.

Now, one of $x-1$ and $x$ is even and the other is odd. Let's say that the even one is a multiple of 4 (since the odd one won't contribute any powers of 2) and the odd one must be a multiple of 25.

So either $x$ is a multiple of 4 and $x-1$ is a multiple of 25, or $x-1$ is a multiple of 4 and $x$ is a multiple of 25. It's not too hard to check all the cases and see that we get $x = 76$ in the first case and $x = 25$ in the second case. When there are more digits, use the Chinese Remainder Theorem to solve the simultaneous congruences.

(Alternatively, $x$ could be a multiple of 4 and of 25, giving $x = 0$, or $x-1$ could be a multiple of 4 and of 25, giving $x = 1$. But these are trivial.)

Call such a number $n$. Then $n^2-n=(n-1)n$ should be divisible by $100\,000=2^5\cdot5^5$. It follows that one of $n-1$ or $n$, call it $n'$, should be divisible by $5^5=3125$, the other by $32$. Therefore we have to find a $k$ with $$n'=k\cdot3125\ \equiv\pm1\qquad({\rm mod}\>32)\ .\tag{1}$$ Now $3125=21$ $({\rm mod}\>32)$, hence $k_*=3$ would satisfy $(1)$ with $=-1$ on the right hand side. But the given size restriction on $n'$ enforces $4\leq k\leq 31$. As $21$ is prime to $32$ there is exactly one $k$ in this range that fulfills $(1)$, namely $k=32-k_*=29$, which then fulfills $(1)$ with $=1$ on the right hand side. This implies that our problem has exactly one solution, namely $n=29 n'=90\,625$.