$(1+\frac{1}{x})\cdot (1+\frac{1}{x+1})\cdots (1+\frac{1}{x+a})=a-x\Leftrightarrow \frac{x+1}{x}\frac{x+2}{x+1}\frac{x+3}{x+2}\cdot ...\frac{x+a}{x+a-1}\cdot \frac{x+a+1}{x+a}=a-x$ $\frac{x+a+1}{x}=a-x\Rightarrow {{x}^{2}}-(a-1)x+1=0$ If both solutions $x$ there are integers, from the product we get that ${{x}_{1}}={{x}_{2}}=1\text{ or }{{x}_{1}}={{x}_{2}}=-1\Rightarrow a\in \left\{ 3,-1 \right\}$ but only $3$ is positive. If ${{x}_{1}}=k\in \mathbb{Z},P=1\Rightarrow {{x}_{2}}=\frac{1}{k}\Rightarrow {{x}_{1}}+{{x}_{2}}=a-1\Leftrightarrow k+\frac{1}{k}=a-1\in \mathbb{N}\Rightarrow k=1\Rightarrow a=3$, and this is the first situation.

I'm sorry but there must be something wrong with your poof because the solutions are $x=2$ and $x=4$ for $a=7$

Can you find the mistake in my solution?

As TuZo above solved, we get that $$\frac{x+a+1}{x}=a-x\implies {{x}^{2}}-(a-1)x+a+1=0$$Now by Vieta's formula, $x_1\cdot x_2=a+1$ and $x_1+x_2=a-1$. Thus we get that both $x_1$ and $x_2$ are integers. Now $(a-1)^{2}-4(a+1)=b^{2}$ must hold, where $b$ is an integer. Hence, $a^{2}-6a-3=(a-3)^{2}-12=b^{2}$. Thus,$$ (a-b-3)(a+b-3)=12=2^{2}\cdot 3$$Solving this diophantine equation, we get that only solution for positive $a$ is $a=7$, so $x_1\cdot x_2=8$ and $x_1+x_2=6$, which implies that $x_1=2$ and $x_2=4$. Answer. The solutions are $x_1=2$ and $x_2=4$ for $a=7$.

TuZo wrote: Can you find the mistake in my solution? Yes TuZo wrote: $\frac{x+a+1}{x}=a-x\Rightarrow {{x}^{2}}-(a-1)x+1=0$ This will be $x^2 -(a-1)x + a +1$

Ok, thank you.

求所有的正整数 $a$, 使得关于 $\mathrm{x}$ 的方程 $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x+1}\right) \ldots\left(1+\frac{1}{x+a}\right)=a-x$ 至少有一个整数解. 并对所有满足条件的正整数 $a$,求相应的整数解. 解：原方程等价于 $\frac{x+a+1}{x}=a-x$ 即$x^2-(a-1)x+a+1=0$ 要使该方程有整数根, 则$\Delta$为完全平方数, 即$a^2-6a-3$为完全平方数. 显然地, 当$a > 7$时,$a^2-8a+16$<$a^2-6a-3$<$a^2-6a+9$,故$a^2-6a-3$不为完全平方数. 因此我们只需考虑$a \ge 7$的情况即可. 枚举知当且仅当$a=7$时,原方程有整数解$x=2,4.$