Let $p$ be the given number and suppose we extend it by $n$ digits. The value of $n$ is still to be determined. The smallest such extension is $p \cdot 10^n$ and the largest is $(p+1) \cdot 10^n - 1$.
It is possible to complete the number $p$ to a perfect square by appending $n$ digits if there exists an integer $m$ such that
$$p \cdot 10^n \leq m^2 < (p+1) \cdot 10^n, \quad \text{ i.e. }$$
$$\sqrt{p} \cdot 10^{\frac{n}{2}} \leq m < \sqrt{p+1} \cdot 10^{\frac{n}{2}}.$$
Because $\sqrt{p+1} - \sqrt{p} > 0$, for $n$ large enough we have $(\sqrt{p+1} - \sqrt{p}) \cdot 10^{\frac{n}{2}} > 1$. Hence, there is an integer between $\sqrt{p} \cdot 10^{\frac{n}{2}}$ and $\sqrt{p+1} \cdot 10^{\frac{n}{2}}$.

Let the given number $p$ have less than $k$ digits, i.e. $p < 10^k$. Let $N$ be the largest integer for which $N^2 \leq p \cdot 10^{k+1}$, i.e.
$$N^2 \leq p \cdot 10^{k+1} < (N+1)^2.$$
Combining $N^2 \leq p \cdot 10^{k+1}$ with $p < 10^k$ we get $N^2 < 10^{2k+1}$, and so $N < \sqrt{10} \cdot 10^k$. Using $1 + 2\sqrt{10} < 10$, which follows from $40 = (2\sqrt{10})^2 < 9^2$, we obtain
$$2N + 1 < 2\sqrt{10} \cdot 10^k + 1 < (1 + 2\sqrt{10}) 10^k < 10^{k+1}.$$
We then get
$$0 < p \cdot 10^{k+1} < (N+1)^2 = N^2 + (2N+1) < p \cdot 10^{k+1} + 10^{k+1} = (p+1) \cdot 10^{k+1}.$$
This means that the number $(N + 1)^2$ is obtained from the number $p$ by appending $k + 1$ digits.