4. Given that $a$, $b$, and $c$ are positive integers, and $a<b<c$. If the product of any two of these numbers minus 1 is divisible by the third number, then $a^{2}+b^{2}+c^{2}=$ $\qquad$
4. 38 .

From the problem, we know
$$
\begin{array}{l}
a \mid (b c - 1), b \mid (a c - 1), c \mid (a b - 1) \\
\Rightarrow a b c \mid (b c - 1)(a c - 1)(a b - 1) \\
\Rightarrow a b c \mid (a b + b c + a c - 1) .
\end{array}
$$

Notice that, $0 < a b + b c + a c - 1 < 3 b c$.
Then $a b + b c + a c - 1 = 2 b c$ or $b c$, which means $a = 1, 2$.
When $a = 1$, $b c \mid (b + b c + c - 1)$, and
$$
b c < b + b c + c - 1 < 2 b c,
$$

which is impossible;
When $a = 2$, $b c \mid (2 b + b c + 2 c - 1)$, and $b c < 2 b + b c + 2 c - 1 < 3 b c$
$$
\Rightarrow 2 b + b c + 2 c - 1 = 2 b c
$$
$$
\begin{array}{l}
\Rightarrow (b - 2)(c - 2) = 3 \\
\Rightarrow (a, b, c) = (2, 3, 5) \\
\Rightarrow a^2 + b^2 + c^2 = 38
\end{array}
$$