Let $m, n, p \in \mathbf{Z}_{+} (m < n < p)$. There are $p$ lamps arranged in a circle, and initially, the pointer points to one of the lamps. Each operation can either rotate the pointer $m$ positions clockwise and then change the on/off state of the lamp it reaches, or rotate the pointer $n$ positions clockwise and then change the on/off state of the lamp it reaches. Find the necessary and sufficient conditions that $m, n, p$ should satisfy so that any of the $2^{p}$ possible on/off states of the lamps can be achieved through these two operations.
For convenience, let's call the lamp initially pointed to by the pointer as lamp 0, and the lamps in the clockwise direction as lamp 1, lamp 2, ..., lamp \( p-1 \). The numbering of the lamps is modulo \( p \), so lamp \( p \) is the same as lamp 0, and so on. The congruence symbol used for the numbering of the lamps is modulo \( p \) by default unless otherwise specified.

Clearly, \( m, n, p \) should be coprime as a whole; otherwise, if there is a common prime factor \( d \), each rotation will be a multiple of this prime factor \( d \), which means that lamps with numbers not multiples of \( d \) will never be reached.

If it is possible to rotate the pointer to any position while keeping the on/off state of all lamps unchanged, then the goal can certainly be achieved. The reason is that after specifying the target state, any lamp that has not reached the target state (let's say it is lamp \( a \)) can be selected, and the pointer can be rotated to lamp \( a-m \) while keeping the on/off state of all lamps unchanged, and then the red knob can be rotated once.

If one of \( m \) or \( n \) is coprime with \( p \), let's assume it is \( m \) (the size relationship between \( m \) and \( n \) does not have a substantial impact), then by rotating the red knob \( p \) times continuously, the on/off state of each lamp will change once, and the pointer will return to its original position. However, as long as the last \( p-1 \) rotations are of the red knob, even if the first rotation is of the blue knob, it will still result in the on/off state of each lamp changing once, but the final position of the pointer will be \( n-m \) positions clockwise from the initial position.

Therefore, if the blue knob is rotated once first, and then the red knob is rotated \( 2p-1 \) times, it will keep the on/off state of all lamps unchanged, and the pointer will rotate \( n-m \) positions.

Thus, in this case, as long as \( n-m \) is coprime with \( p \), the condition is met.

If both \( m \) and \( n \) are not coprime with \( p \), but \( n-m \) is coprime with \( p \), let \( (m, p) = s \), then the above argument still holds if \( p \) is replaced with \( \frac{p}{s} \). 2015, . 011
The following discussion is based on \( (n-m, p) \) (denoted as \( k \)).
Lamps with numbers congruent to \( im \) modulo \( k \) are called the \( i \)-th group (\( i=0,1, \cdots, k-1 \)). Then, starting from the \( i \)-th group (\( i=0,1, \cdots, k-2 \)), whether rotating \( m \) positions clockwise or \( n \) positions clockwise, it will always reach the \( (i+1) \)-th group; and starting from the \( (k-1) \)-th group, whether rotating \( m \) positions clockwise or \( n \) positions clockwise, it will always reach the 0-th group. It is evident that the group numbering of the lamps is understood modulo \( k \).

Therefore, starting from the initial position, no matter how the operations are performed, the lamps whose on/off state changes will always follow the order of the 1st group, the 2nd group, ..., the \( (k-1) \)-th group, and the 0-th group in a cycle.

Assign the value 1 to the off lamps and -1 to the on lamps, and let the product of the values of all lamps in the \( i \)-th group be \( P_i \). Then, initially, all \( P_i \) are equal to 1, and each operation changes the sign of \( P_1, P_2, \cdots, P_{k-1}, P_0 \) in a cyclic order, and after each sign change, all 1s in \( P_1, P_2, \cdots, P_{k-1}, P_0 \) are adjacent, and all -1s are also adjacent.

If \( k > 3 \), then the state where \( P_2 = -1 \) and \( P_i = 1 \) (\( i \neq 2 \)) can never be reached, so it does not meet the requirements.

If \( k = 2 \), all lamps are divided into two groups, and since \( m, n, p \) are coprime as a whole, \( m \) and \( n \) are both odd, and \( p \) is even.

According to a similar reason as before, the pointer can be rotated to any position within the same group while keeping the on/off state of all lamps unchanged. Therefore, first, select a lamp in the 1st group (i.e., all lamps with odd numbers) that does not match the target on/off state, and rotate the pointer to the position \( m \) positions before it (which is an even number), and then rotate to that position. Next, use a similar method to handle a lamp in the 0th group that does not match the target on/off state, to reduce the number of lamps in the two groups that do not match the target on/off state by 1 alternately. If at some point, the group to be rotated to already has all lamps matching the target on/off state, just rotate once to make the number of lamps in that group that do not match the target on/off state 1, and then change it back to 0 the next time. Repeating this process, the number of lamps in both groups that do not match the target on/off state can eventually be reduced to 0.

In summary, the necessary and sufficient condition is that \( m, n, p \) are coprime as a whole and \( (n-m, p) \leq 2 \), or equivalently, \( (n-m, p) \leq 2 \) and \( m, n, p \) are not all even.