Positive Integers whose Square Root equals Sum of Digits

Theorem
The following positive integers have a square root that equals the sum of their digits:
 * $0, 1, 81$

and there are no more.

Proof
By considering the square roots, we are looking for positive integers with a square for which its sum of digits is equal to the original number.

It is easy to verify the result up to $36$.

We prove that for $n > 36$, the sum of digits of $n^2$ cannot equal $n$.

Let $n$ be a $d$-digit number.

Then $n < 10^d$.

Thus $n^2 < 10^{2 d}$.

Therefore $n^2$ has at most $2 d$ digits.

Hence the sum of digits of $n^2$ is at most $18 d$.

If $n$ is a $2$-digit number, we have $n > 36 = 18 d$.

Now suppose $d \ge 3$.

Then:

Therefore for $n > 36$, the sum of digits of $n^2$ cannot equal $n$.