4 Find all positive integer triples $(a, b, c)$ that satisfy $a^{2}+b^{2}+c^{2}=2005$, and $a \leqslant b \leqslant c$. (Provided by Tao Pingsheng)
Since any odd square number leaves a remainder of 1 when divided by 4, and any even square number is a multiple of 4, and since 2005 leaves a remainder of 1 when divided by 4, it follows that among the three numbers \(a^2, b^2, c^2\), there must be two even squares and one odd square.

Let \(a = 2m, b = 2n, c = 2k - 1\), where \(m, n, k\) are positive integers. The original equation can be transformed into:
\[
m^2 + n^2 + k(k-1) = 501.
\]

Since the remainder of any square number when 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 \(m\) and \(n\) are both multiples of 3.
Let \(m = 3m_1, n = 3n_1\), and \(\frac{k(k-1)}{3}\) is an integer. From (1),
\[
3m_1^2 + 3n_1^2 + \frac{k(k-1)}{3} = 167.
\]

Thus, \(\frac{k(k-1)}{3} \equiv 167 \equiv 2 \pmod{3}\).
Let \(\frac{k(k-1)}{3} = 3r + 2\), then
\[
k(k-1) = 9r + 6.
\]

And from (1), \(k(k-1) < 501\), so \(k \leq 22\).
Hence, from (3), \(k\) can take the values \(3, 7, 12, 16, 21\), and substituting into (2) we get the following cases:
\[
\begin{array}{c}
\left\{\begin{array}{l}
k = 3, \\
m_1^2 + n_1^2 = 55,
\end{array}\right. \quad \left\{\begin{array}{l}
k = 7, \\
m_1^2 + n_1^2 = 51,
\end{array}\right. \quad \left\{\begin{array}{l}
k = 12, \\
m_1^2 + n_1^2 = 41,
\end{array}\right. \\
\left\{\begin{array}{l}
k = 16, \\
m_1^2 + n_1^2 = 29,
\end{array}\right. \quad \left\{\begin{array}{l}
k = 21, \\
m_1^2 + n_1^2 = 9.
\end{array}\right.
\end{array}
\]

Since \(55\) and \(51\) are of the form \(4N + 3\) and cannot be expressed as the sum of two squares, and \(9\) cannot be expressed as the sum of two positive integers' squares, the only solutions are when \(k = 12\) and \(k = 16\).

When \(k = 12\), from \(m_1^2 + n_1^2 = 41\), we get \((m_1, n_1) = (4, 5)\), then \(a = 6m_1 = 24, b = 6n_1 = 30, c = 2k - 1 = 23\), so \((a, b, c) = (24, 30, 23)\).

When \(k = 16\), from \(m_1^2 + n_1^2 = 29\), we get \((m_1, n_1) = (2, 5)\), then \(a = 6m_1 = 12, b = 6n_1 = 30, c = 2k - 1 = 31\), so \((a, b, c) = (12, 30, 31)\).

(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 (1) to discuss each case:
If \(k = 2\), then \(m_1^2 + n_1^2 = 499\), and \(499 \equiv 3 \pmod{4}\), so there is no solution.
If \(k = 5\), then \(m_1^2 + n_1^2 = 481\), using the identity
\[
\begin{aligned}
\left(\alpha^2 + \beta^2\right)\left(x^2 + y^2\right) & = (\alpha x + \beta y)^2 + (\alpha y - \beta x)^2 \\
& = (\alpha x - \beta y)^2 + (\alpha y + \beta x)^2,
\end{aligned}
\]
we have
\[
\begin{aligned}
481 & = 13 \times 37 = \left(3^2 + 2^2\right)\left(6^2 + 1^2\right) \\
& = 20^2 + 9^2 = 16^2 + 15^2.
\end{aligned}
\]

So \((m, n) = (9, 20)\) or \((15, 16)\).
Thus, we get two solutions \((a, b, c) = (2m, 2n, 2k-1) = (18, 40, 9)\) 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 \((11, 18)\), giving two solutions \((a, b, c) = (2m, 2n, 2k-1) = (4, 42, 15)\) or \((22, 36, 15)\).
If \(k = 11\), then \(m_1^2 + n_1^2 = 391\), and \(391 \equiv 3 \pmod{4}\), so there is no solution.
If \(k = 14\), then \(m_1^2 + n_1^2 = 319\), and \(319 \equiv 3 \pmod{4}\), so there is no solution.
If \(k = 17\), then \(m_1^2 + n_1^2 = 229\), and \(229 = 15^2 + 2^2\), so \((m, n) = (2, 15)\), giving one solution \((a, b, c) = (2m, 2n, 2k-1) = (4, 30, 33)\).

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, so there are 7 solutions in total: \((23, 24, 30), (12, 30, 31), (9, 18, 40), (9, 30, 32), (4, 15, 42), (15, 22, 36), (4, 30, 33)\).
Upon verification, all of them satisfy the equation.