Determine all positive integers $a$ for which the equation $$\left ( 1 + \frac {1} {x} \right ) \cdot \left ( 1 + \frac {1} {x+1} \right ) \cdots \left ( 1 + \frac {1} {x+a} \right ) = a - x $$ has at least one integer solution $x$.
For each such integer $a$, determine the corresponding solutions.
On
The LHS can be rewritten as
$$\left ( 1 + \frac {1} {x} \right ) \cdot \left ( 1 + \frac {1} {x+1} \right ) \cdots \left ( 1 + \frac {1} {x+a} \right ) = \frac{(x+1)\cdots(x+a+1)}{x(x+1)\cdots(x+a)} = \frac{x+a+1}{x} $$
Hence, the problem reduces to $$ \frac{x+a+1}{x} = a-x $$ Since we are lookign for integer value of $a$ and $x$, the RHS is an integer. Thus, $\frac{x+a+1}{x}$ must also be an integer. Therefore, $x | a+1$.
Moving on, we have that $$ x+a+1 = (a-x)x \iff x^2-(a-1)x+(a+1) = 0 $$
Can you continue from here?
P.S.: The real solutions are
$$x_1 = \frac{a-1 - \sqrt{a^2 -6a -3}}{2} \quad \text{and} \quad x_2 = \frac{a-1 +\sqrt{a^2 -6a -3}}{2} $$
On
Clearly the LHS of the equation simplifies to $\frac{x+a+1}{x}$. But note that, this also enforces that our solution (say $x'$) to the equation $$\frac{x+a+1}{x} = a- x$$ must satisfy $x'\neq0, x'\neq -1, \ldots, x' \neq -a$.
Now, the equation at hand is $x^2 + (1-a)x +a+1 = 0$, for which we need to find the positive integers $a$ such that the equation has at least one integer solution.
Considering the determinant of the equation, $D = (a-3)^2 - 12 \geq 0$, for the equation to have real solutions. Also, $a>0$. Thus, $a \in [3 + 2\sqrt3, \infty)$.
The solutions must be an integer, which will be true if the discriminant $D$ is a perfect square.
Hence, $(a-3)^2 - 12$ is a perfect square. We can change this to the following equation: $$(a-3)^2 - 12 = b^2$$ for some integers $a,b$ where $a \in [3 + 2\sqrt3, \infty)$ and as $a$ is an integer, we have $a \in [7, \infty)$.
The above equation gives us $(a-3+b)(a-3-b) = 12$. Now, we can consider the pair of integers whose product gives us 12. Note that, both the integers should be odd, or both should be even.
This gives us the following pairs: (2, 6), (6, 2), (-2, -6), (-6, -2).
Solving for
Clearly, only the first two give us a valid pair as $a\in [7, \infty)$. Hence, $a=7$. Now, we check the solutions of the quadratic equation, which becomes $$x^2 - 6x +8 = 0$$ The roots are $x' = 4$ and $x' = 2$, which also satisfy the enforced inequalities, which we stated in the beginning.
The terms of the product can be made into simple fractions as $$1+\frac{1}{x+k}=\frac{x+k+1}{x+k},$$ from which it is clear that the numerator of each term cancels against the denominator of the next term, so \begin{eqnarray*} a-x&=&\left ( 1 + \frac {1} {x} \right ) \cdot \left ( 1 + \frac {1} {x+1} \right ) \cdots \left ( 1 + \frac {1} {x+a} \right )\\ &=&\left (\frac {x+1} {x} \right ) \cdot \left ( \frac {x+2} {x+1} \right ) \cdots \left ( \frac {x+a+1} {x+a} \right )\\ &=&\frac{x+a+1}{x}. \end{eqnarray*} Then clearing the denominator and rearranging shows that $x$ is a root of the quadratic $$X^2+(1-a)X+(1+a)=0,$$ which has an integral root if and only if its discriminant is a perfect square, which is $$\Delta=(1-a)^2-4(1+a)=a^2-6a-3=(a-3)^2-12.$$ Of course if $a$ is too large then $(a-3)^2-12$ cannot be a perfect square; if $a>9$ then $$(a-3)^2>\Delta=(a-3)^2-12=a^2-6a-3>a^2-8a+16=(a-4)^2,$$ and so $a\leq9$. It is then a quick check to find that only $a=7$ works, corresponding to $x=2$ and $x=4$.