Let $a$ be a positive integer. Determine all $a$ such that the equation
$$ \biggl( 1+\frac{1}{x} \biggr) \cdot \biggl( 1+\frac{1}{x+1} \biggr) \cdots \biggl( 1+\frac{1}{x+a} \biggr)=a-x$$
has at least one integer solution for $x$.
For every such $a$ state the respective solutions.

(Richard Henner)
1. We start with the given equation:
   \[
   \left( 1 + \frac{1}{x} \right) \left( 1 + \frac{1}{x+1} \right) \cdots \left( 1 + \frac{1}{x+a} \right) = a - x
   \]

2. We simplify the left-hand side of the equation. Each term in the product can be rewritten as:
   \[
   1 + \frac{1}{x+k} = \frac{x+k+1}{x+k}
   \]
   Therefore, the product becomes:
   \[
   \prod_{k=0}^{a} \left( 1 + \frac{1}{x+k} \right) = \prod_{k=0}^{a} \frac{x+k+1}{x+k}
   \]

3. Notice that this product is a telescoping product:
   \[
   \frac{x+1}{x} \cdot \frac{x+2}{x+1} \cdot \frac{x+3}{x+2} \cdots \frac{x+a+1}{x+a} = \frac{x+a+1}{x}
   \]

4. Substituting this back into the original equation, we get:
   \[
   \frac{x+a+1}{x} = a - x
   \]

5. Multiplying both sides by \(x\) to clear the fraction:
   \[
   x + a + 1 = x(a - x)
   \]

6. Rearranging terms to form a quadratic equation:
   \[
   x^2 - (a-1)x + (a+1) = 0
   \]

7. Using Vieta's formulas, the sum and product of the roots \(x_1\) and \(x_2\) of the quadratic equation \(x^2 - (a-1)x + (a+1) = 0\) are:
   \[
   x_1 + x_2 = a - 1
   \]
   \[
   x_1 x_2 = a + 1
   \]

8. For \(x_1\) and \(x_2\) to be integers, the discriminant of the quadratic equation must be a perfect square:
   \[
   (a-1)^2 - 4(a+1) = b^2
   \]
   Simplifying, we get:
   \[
   a^2 - 6a - 3 = b^2
   \]

9. This can be rewritten as:
   \[
   (a - 3)^2 - b^2 = 12
   \]
   This is a difference of squares:
   \[
   (a - 3 - b)(a - 3 + b) = 12
   \]

10. We solve the Diophantine equation by considering the factor pairs of 12:
    \[
    (a - 3 - b)(a - 3 + b) = 12
    \]
    The factor pairs of 12 are: \((1, 12), (2, 6), (3, 4), (-1, -12), (-2, -6), (-3, -4)\).

11. Solving for each pair:
    - For \((1, 12)\):
      \[
      a - 3 - b = 1 \quad \text{and} \quad a - 3 + b = 12
      \]
      Adding these equations:
      \[
      2(a - 3) = 13 \implies a - 3 = \frac{13}{2} \quad \text{(not an integer)}
      \]

    - For \((2, 6)\):
      \[
      a - 3 - b = 2 \quad \text{and} \quad a - 3 + b = 6
      \]
      Adding these equations:
      \[
      2(a - 3) = 8 \implies a - 3 = 4 \implies a = 7
      \]
      Subtracting these equations:
      \[
      2b = 4 \implies b = 2
      \]

    - For \((3, 4)\):
      \[
      a - 3 - b = 3 \quad \text{and} \quad a - 3 + b = 4
      \]
      Adding these equations:
      \[
      2(a - 3) = 7 \implies a - 3 = \frac{7}{2} \quad \text{(not an integer)}
      \]

    - The negative pairs do not yield positive \(a\).

12. Thus, the only solution for positive \(a\) is \(a = 7\).

13. Substituting \(a = 7\) back into the quadratic equation:
    \[
    x^2 - 6x + 8 = 0
    \]
    Solving this quadratic equation:
    \[
    x = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2}
    \]
    \[
    x = 4 \quad \text{or} \quad x = 2
    \]

The final answer is \( \boxed{ x_1 = 2 } \) and \(x_2 = 4\) for \(a = 7\).