## Task 14/68

We are looking for all natural numbers that are equal to the square of their cross sum.
If $q$ is the cross sum of the sought number $x$, then according to the problem statement, the equation $x=q^{2}$ should hold. Now, $q$ and $q^{2}$ belong to the same residue class $\bmod 9$:

$$
q^{2} \equiv q(\bmod 9) \rightarrow q(q-1) \equiv 0(\bmod 9)
$$

which yields $q \equiv 0(\bmod 9)$ or $q \equiv 1(\bmod 9)$.

The single-digit $q$-values 1 and 9 satisfy the conditions of the problem: $1=1^{2}$ and $81=9^{2}$.

If the cross sum has $n$ digits $(n \geq 2)$, then $10^{n-1} \leq q_{n} \leq 10^{n}(2)$, so $q_{n}^{2}$ has either $2 n-1$ or $2 n$ digits. The cross sum of such a number is at most $9 \cdot 2 n=18 n$. Therefore, $q_{n} \leq 18 n$ (3).

From inequalities (2) and (3), it follows as a necessary condition that $10^{n-1} \leq 18 n$. (4)

This inequality is satisfied for the natural numbers 1 and 2. However, these are also the only natural numbers for which it holds; as one can easily see from a graphical representation of the functions $f(x)=10^{x-1}$ and $g(x)=18 x$. For $n=3$, this inequality no longer holds.

We now compile the two-digit $q$-values and their squares that, taking into account inequality (3), satisfy the congruences (1):

$$
\begin{array}{r}
q=10 ; 18 ; 19 ; 27 ; 28 ; 36 \\
q^{2}=100 ; 324 ; 361 ; 729 ; 784 ; 1296
\end{array}
$$

It turns out that the cross sums of $q^{2}$ do not yield $q$, but are only congruent to $q$ (mod 9). Therefore, the problem has only the two solutions $x=1$ and $x=81$ (where $x=1$ can be considered trivial).