2. (India 2004) $S$ is the set of all ordered tuples $(a, b, c, d, e, f)$ where $a, b, c, d, e, f$ are integers, and $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=f^{2}$. Find the largest $k$ such that for all elements of $S$, $k$ divides $a b c d e f$.


2. $a=b=c=1, d=2, e=3, f=4$. Then $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=f^{2}$. And abcdef $=24$. It follows that $k \leqslant 24$.

Below is the proof: $k=24$. First, among $a b c$ def, there must be an odd number, otherwise, if we take both sides modulo 8, the left side will have a remainder of 5, and the right side will have a remainder of 1, which is a contradiction.
(1) If $f$ is even, then $a b c d e$ must have an even number. Suppose $e$ is even, then
$$
a^{2}+b^{2}+c^{2}+d^{2}=4\left(m^{2}-k^{2}\right),
$$

where $f=2 m, e=2 k$.
If $a b c d$ are odd, then modulo 8, the left side will have a remainder of 4, so $m^{2}-k^{2}$ must be odd, which means $m, k$ are one odd and one even, so $e f$ is a multiple of 8.
If $a b c d$ has an even number, then $a b c d e f$ is divisible by 8.
(2) If $e$ is even, then $a b c d$ must have an even number, otherwise, the right side modulo 8 will have a remainder of 1, while the left side modulo 8 will have a remainder of 0 or 4. So, without loss of generality, assume $d, e$ are even, then both sides modulo 4, the right side will have a remainder of 1, while the left side $\equiv a^{2}+b^{2}+c^{2}$, so $a, b, c$ are all odd, then the right side $\equiv 3(\bmod 4)$, which is a contradiction. Therefore, $a, b, c$ must have an even number, so $8 \mid a b c d e f$.
Below is the proof: among abcdef, there must be a multiple of 3, otherwise $x^{2} \equiv 1(\bmod 3)$, both sides modulo 3, $2 \equiv 1(\bmod 3)$, which is a contradiction.
So $3 \mid a b c d e f$.
So $24 \mid a b c d e f$.
So $k_{\max }=24$.