26. (2004 Western China Mathematical Olympiad) Find all positive integer triples \((a, b, c)\) satisfying \(a^{2}+b^{2}+c^{2}=2005\), and \(a \leqslant b \leqslant c\).
26. Since any odd square number divided by 4 leaves a remainder of 1, and any even square number is a multiple of 4, and since 2005 divided by 4 leaves a remainder of 1, it follows that among $a^{2}$, $b^{2}$, $c^{2}$, there must be two even square numbers and one odd square number.
Let $a=2 m, b=2 n, c=2 k-1, m, n, k$ be positive integers, and the original equation can be transformed into:
$m^{2}+n^{2}+k(k-1)=501$.
Also, because the remainder of any square number divided by 3 is either 0 or 1, we now discuss $k$:
(i) If $3 \mid k(k-1)$, then from (1), $3 \mid m^{2}+n^{2}$, so $m, n$ are both multiples of 3.
Let $m=3 m_{1}, n=3 n_{1}$, and $\frac{k(k-1)}{3}$ is an integer, from (1)
$$
3 m_{1}^{2}+3 n_{1}^{2}+\frac{k(k-1)}{3}=167,
$$

thus $\frac{k(k-1)}{3} \equiv 167 \equiv 2(\bmod 3)$.
Let $\frac{k(k-1)}{3}=3 r+2$, then
$$
k(k-1)=9 r+6 \text {. }
$$

and from (1), $k(k-1)<501$, so $k \leqslant 22$.
Hence, from (3), $k$ can take the values $3,7,12,16,21$, substituting into (2) respectively gives the following cases:
$$
\left\{\begin{array} { l } 
{ k = 3 , } \\
{ m _ { 1 } ^ { 2 } + n _ { 1 } ^ { 2 } = 5 5 , }
\end{array} \quad \left\{\begin{array}{l}
k=7, \\
m_{1}^{2}+n_{1}^{2}=51,
\end{array},\left\{\begin{array} { l } 
{ k = 1 2 , } \\
{ m _ { 1 } ^ { 2 } + n _ { 1 } ^ { 2 } = 4 1 , }
\end{array} \left\{\begin{array} { l } 
{ k = 1 6 , } \\
{ m _ { 1 } ^ { 2 } + n _ { 1 } ^ { 2 } = 2 9 , }
\end{array} \quad \left\{\begin{array}{l}
k=21, \\
m_{1}^{2}+n_{1}^{2}=9 .
\end{array}\right.\right.\right.\right.\right.
$$

Since 55 and 51 are both of the form $4 N+3$, they cannot be expressed as the sum of two squares, and 9 cannot be expressed as the sum of two positive integers' squares, thus there are positive integer solutions $m_{1}, n_{1}$ only when $k=12$ and $k=16$.

When $k=12$, from $m_{1}^{2}+n_{1}^{2}=41$, we get $\left(m_{1}, n_{1}\right)=(4,5)$, then $a=6 m_{1}=24, b=6 n_{1}=30, c=2 k-1=$ 23, thus $(a, b, c)=(24,30,23)$.

When $k=16$, from $m_{1}^{2}+n_{1}^{2}=29$, we get $\left(m_{1}, n_{1}\right)=(2,5)$, then $a=6 m_{1}=12, b=6 n_{1}=30, c=2 k-1=$ 31, thus $(a, b, c)=(12,30,31)$.
(ii) If $3 \nmid k(k-1)$, since any three consecutive numbers must include one multiple of 3, then $k+1$ is a multiple of 3, so $k$ leaves a remainder of 2 when divided by 3, thus $k$ can only take the values $2,5,8,11,14,17,20$.
Using equation (1) to discuss each case respectively:
If $k=2$, then $m_{1}^{2}+n_{1}^{2}=499$, and $499 \equiv 3(\bmod 4)$, at this time there is no solution.
If $k=5$, then $m_{1}^{2}+n_{1}^{2}=481$, using the identity
$$
\left(\alpha^{2}+\beta^{2}\right)\left(x^{2}+y^{2}\right)=(\alpha x+\beta y)^{2}+(\alpha y-\beta x)^{2}=(\alpha x-\beta y)^{2}+(\alpha y+\beta x)^{2},
$$

we know $481=13 \times 37=\left(3^{2}+2^{2}\right)\left(6^{2}+1^{2}\right)=20^{2}+9^{2}=16^{2}+15^{2}$.
So $(m, n)=(9,20)$ or $(15,16)$.
Thus we get two solutions $(a, b, c)=(2 m, 2 n, 2 k-1)=(18,40,9)$ or $(30,32,9)$.
If $k=8$, then $m_{1}^{2}+n_{1}^{2}=445$, and $445=5 \times 89=\left(2^{2}+1^{2}\right)\left(8^{2}+5^{2}\right)=21^{2}+2^{2}=18^{2}+11^{2}$, so, $(m, n)=$ $(2,21)$ or $(11,18)$, we get two solutions $(a, b, c)=(2 m, 2 n, 2 k-1)=(4,42,15)$ or $(22,36,15)$.
If $k=11$, then $m_{1}^{2}+n_{1}^{2}=391$, and $391 \equiv 3(\bmod 4)$, at this time there is no solution.
If $k=14$, then $m_{1}^{2}+n_{1}^{2}=319$, and $319 \equiv 3(\bmod 4)$, at this time there is no solution.
If $k=17$, then $m_{1}^{2}+n_{1}^{2}=229$, and $229=15^{2}+2^{2}$, so $(m, n)=(2,15)$, we get one solution $(a, b, c)=$ $(2 m, 2 n, 2 k-1)=(4,30,33)$.

If $k=20$, then $m_{1}^{2}+n_{1}^{2}=121=11^{2}$, and $11^{2}$ cannot be expressed as the sum of two positive integers' squares, thus this problem has 7 solutions: $(23,24,30),(12,30,31),(9,18,40),(9,30,32),(4,15,42),(15,22,36),(4,30,33)$.
Upon verification, they all satisfy the equation.