10.58 Find all natural numbers $n$, such that their unit digit is 4, and the sum of the squares of their digits is not less than $n$.
(Leningrad Mathematical Olympiad, 1980)
[Solution] Let $n=\overline{a_{k} a_{k-1} \cdots a_{2} a_{1} 4}$, where $a_{i} \in\{0,1,2, \cdots, 9\}, \quad a_{k} \neq 0$. By the problem statement, we have
$$
a_{k}^{2}+a_{k-1}^{2}+\cdots+a_{1}^{2}+16 \geqslant n>a_{k} \cdot 10^{k} .
$$

From this, we can derive

that is,
$$
\begin{array}{l}
k \cdot 9^{2}+16>a_{k} \cdot 10^{k} \geqslant 10^{k}, \\
81 k+16>10^{k} .
\end{array}
$$

When $k \geqslant 3$,
$$
\begin{array}{c}
10^{k}=(9+1)^{k}>9^{k}+k \cdot 9^{k-1} \\
>16+k \cdot 9^{2} \\
k \leqslant 2 .
\end{array}
$$

Thus,
when $k=0$, $n=4$, in this case $4^{2}=16>4=n$, which meets the problem's requirement.
When $k=1$, let $n=10 x+4$, then by the problem statement,
$$
\begin{array}{r}
x^{2}+16>10 x+4 \\
(x-5)^{2} \geqslant 13 .
\end{array}
$$

Clearly, only when $x=1$ and $x=9$ satisfy the requirement, at this time, $n=14$ and $n=94$.
When $k=2$, let $n=100 x+10 y+4$, then by the problem statement,
$$
\begin{array}{c}
x^{2}+y^{2}+16 \geqslant 100 x+10 y+4, \\
(50-x)^{2}+(y-5)^{2} \geqslant 50^{2}+5^{2}-12 .
\end{array}
$$

Since
$$
49^{2} \geqslant(50-x)^{2}, \quad 5^{2} \geqslant(y-5)^{2},
$$

then
$$
\begin{array}{c}
49^{2}+5^{2} \geqslant(50-x)^{2}+(y-5)^{2} \geqslant 50^{2}+5^{2}-12 \\
12 \geqslant 50^{2}-49^{2}=99
\end{array}
$$

a contradiction arises.
Therefore, the solutions for $n$ are only $n=4,14,94$ three solutions.