Let $k$ be the smallest positive integer for which there exist distinct integers $m_{1}, m_{2}, m_{3}, m_{4}, m_{5}$ such that the polynomial $$p(x)=\left(x-m_{1}\right)\left(x-m_{2}\right)\left(x-m_{3}\right)\left(x-m_{4}\right)\left(x-m_{5}\right)$$ has exactly $k$ nonzero coefficients. Find, with proof, a set of integers $m_{1}, m_{2}, m_{3}, m_{4}, m_{5}$ for which this minimum $k$ is achieved.
1. **Claim:** \( k \geq 3 \)

2. **Proof:**
   - For \( k = 1 \), the polynomial would be \( x^5 = 0 \), which implies that all roots are zero, repeated five times. This is not possible since the roots must be distinct.
   - For \( k = 2 \), the polynomial would be of the form \( x^5 - rx^n = 0 \) for some \( n \leq 5 \). To have five distinct real roots, \( n \) must be at least 1. If \( n = 1 \), the polynomial becomes \( x^4 - r = 0 \), which factors as \( (x^2 + \sqrt{r})(x^2 - \sqrt{r}) = 0 \). However, this does not yield five distinct roots, so \( k = 2 \) is also not possible.

3. **Construction for \( k = 3 \):**
   - Consider the polynomial \( (x-0)(x-1)(x+1)(x-3)(x+3) \).
   - Expanding this polynomial, we get:
     \[
     (x-0)(x-1)(x+1)(x-3)(x+3) = x(x^2 - 1)(x^2 - 9) = x(x^4 - 10x^2 + 9) = x^5 - 10x^3 + 9x
     \]
   - This polynomial has exactly three nonzero coefficients: \( x^5 \), \( -10x^3 \), and \( 9x \).

4. **General Case:**
   - Given \( n \geq 1 \), we need to find the smallest integer \( k \) such that the polynomial \( (x - m_1)(x - m_2) \cdots (x - m_n) \) has exactly \( k \) nonzero coefficients with distinct integer roots \( m_1, m_2, \ldots, m_n \).
   - By Descartes' Rule of Signs, a polynomial with \( k \) nonzero coefficients can have at most \( k-1 \) positive roots and \( k-1 \) negative roots.
   - Therefore, the total number of distinct roots is at most \( (k-1) + (k-1) + 1 = 2k - 1 \) (considering zero as a root).
   - By the Fundamental Theorem of Algebra, the polynomial has \( n \) distinct integer roots. Hence, \( n \leq 2k - 1 \).
   - Solving for \( k \), we get:
     \[
     \frac{n + 1}{2} \leq k \implies k \geq \left\lfloor \frac{n}{2} \right\rfloor + 1
     \]
   - Therefore, the minimum \( k \) is \( \left\lfloor \frac{n}{2} \right\rfloor + 1 \).

5. **Construction:**
   - For \( n = 2k - 2 \), the polynomial can be constructed as:
     \[
     P(x) = (x + 1)(x - 1)(x + 2)(x - 2) \cdots (x + (k-1))(x - (k-1))
     \]
   - For \( n = 2k - 1 \), the polynomial can be constructed as:
     \[
     P(x) = x(x + 1)(x - 1)(x + 2)(x - 2) \cdots (x + (k-1))(x - (k-1))
     \]

\(\blacksquare\)

The final answer is \( \boxed{ k = 3 } \)