18.3.19 $\star \star$ Find all positive integer triples $(a, b, c)$ that satisfy $a^{2}+b^{2}+c^{2}=2005$ and $a \leqslant b \leqslant c$.
Given $44<\sqrt{2005}<45,25<\sqrt{\frac{2005}{3}}<26$, so
$$
26 \leqslant c \leqslant 44 .
$$
$a^{2}+b^{2}=0(\bmod 3)$, if and only if $a=b=0(\bmod 3)$. Therefore, $3 \mid a^{2}+b^{2}$ implies $9 \mid a^{2}+b^{2}$.

If $3 \nmid c$, then $c^{2} \equiv 1(\bmod 3), a^{2}+b^{2}=2005-c^{2} \equiv 0(\bmod 3)$, thus $9 \mid a^{2}+b^{2}, c^{2} \equiv 2005 \equiv 7(\bmod 9), c \equiv 4,5(\bmod 9)$.
Among (1), the only values of $c$ that satisfy $3 \mid c$ or $c \equiv 4,5(\bmod 9)$ are
$$
\begin{aligned}
c & =27,30,33,36,39,42,31,32,40,41 . \\
2005-27^{2} & =1276=4 \times 319, \\
2005-39^{2} & =484=4 \times 121, \\
2005-41^{2} & =324=4 \times 81,
\end{aligned}
$$
$319,121,81$ cannot be expressed as the sum of two positive integers squared $\left(319 \equiv 3 \neq a^{2}+b^{2}(\bmod 4) .121\right.$ and 81 can be easily verified). And $4 \mid a^{2}+b^{2}$ implies $a, b$ are both even. But $\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}$ cannot equal $319,121,81$, so $a^{2}+b^{2} \neq 4 \times 319,4 \times 121,4 \times 51$.
$$
\begin{array}{l}
2005-30^{2}=1105=529+576=23^{2}+24^{2}, \\
2005-31^{2}=1044=144+900=12^{2}+30^{2}, \\
2005-32^{2}=981=81+900=9^{2}+30^{2}, \\
2005-33^{2}=916=16+900=4^{2}+30^{2}, \\
2005-36^{2}=709=225+484=15^{2}+22^{2}, \\
2005-40^{2}=405=81+324=9^{2}+18^{2}, \\
2005-42^{2}=241=16+225=4^{2}+15^{2} .
\end{array}
$$

Thus, there are 7 solutions to this problem, which are
$$
(a, b, c)=(23,24,30),(12,30,31),(9,30,32),(4,30,33),(15,22, 
$$