Consider the function $$f(x)=\sum_{k=1}^{m}(x-k)^{4}~, \qquad~ x \in \mathbb{R}$$ where $m>1$ is an integer. Show that $f$ has a unique minimum and find the point where the minimum is attained.

1. **Existence of Minima:**
   To determine if the function \( f(x) = \sum_{k=1}^{m}(x-k)^{4} \) has a minimum, we first compute the second derivative \( f''(x) \). 

   \[
   f(x) = \sum_{k=1}^{m}(x-k)^{4}
   \]

   The first derivative \( f'(x) \) is:

   \[
   f'(x) = \sum_{k=1}^{m} 4(x-k)^{3}
   \]

   The second derivative \( f''(x) \) is:

   \[
   f''(x) = \sum_{k=1}^{m} 12(x-k)^{2}
   \]

   Since \( (x-k)^{2} \geq 0 \) for all \( x \in \mathbb{R} \) and \( k \in \{1, 2, \ldots, m\} \), it follows that \( f''(x) \geq 0 \). In fact, \( f''(x) > 0 \) for all \( x \in \mathbb{R} \) because the sum of non-negative terms is positive unless all terms are zero, which is not the case here. Therefore, \( f(x) \) is strictly convex, implying that \( f(x) \) has a unique minimum.

2. **Uniqueness of Minima:**
   Since \( f''(x) > 0 \) for all \( x \in \mathbb{R} \), \( f(x) \) is strictly convex. A strictly convex function has at most one critical point, and this critical point is where the minimum occurs. Thus, \( f(x) \) has a unique minimum.

3. **Finding the Minimum:**
   To find the point where the minimum is attained, we set the first derivative \( f'(x) \) to zero and solve for \( x \):

   \[
   f'(x) = 4 \sum_{k=1}^{m} (x-k)^{3} = 0
   \]

   We can simplify this by considering the symmetry of the function. Notice that the function \( f(x) \) is symmetric around the midpoint of the interval \([1, m]\). The midpoint is:

   \[
   x = \frac{1 + m}{2}
   \]

   To verify, we substitute \( x = \frac{m+1}{2} \) into \( f'(x) \):

   \[
   f'\left(\frac{m+1}{2}\right) = 4 \sum_{k=1}^{m} \left(\frac{m+1}{2} - k\right)^{3}
   \]

   By symmetry, for each term \( \left(\frac{m+1}{2} - k\right)^{3} \), there is a corresponding term \( \left(k - \frac{m+1}{2}\right)^{3} \) such that their sum is zero. Therefore, the sum of all these terms is zero, confirming that:

   \[
   f'\left(\frac{m+1}{2}\right) = 0
   \]

   Hence, the unique minimum of \( f(x) \) is attained at \( x = \frac{m+1}{2} \).

The final answer is \( \boxed{\frac{m+1}{2}} \)