Find all positive primes $p_1,p_2,p_3, \cdots p_n$ such that $$\left(1+\frac{1}{p_1}\right)\left(1+\frac{1}{p_2}\right) \cdots \left(1+\frac{1}{p_n}\right) =2$$
I found this question while finding all squarefree perfect
numbers.I denoted the primes as $p_1,p_2, \ldots, p_n$ thus the sum of all the factors is $$1+p_1+p_2+ \cdots p_1 p_2+\cdots +p_1 p_2 \cdots p_n$$ which we can notice to be $$\alpha =(1+p_1)(1+p_2) \cdots (1+p_n)$$
Thus, $$\alpha=2 p_1 p_2 \cdots p_n$$
Now dividing(transposing) each $(1+p_k)$ by $p_k$ we get the proposed question.
I think there must be some cancellation in the fractions thus we assume WLOG they are in ascending order but still we don't know which factors cancelled where 🙁
Best Answer
Let $p_n=\max\{p_1,...,p_n\}$.
Thus, $1+p_i$ for some $i$ is divisible by $p_n$, which is possible only for $n=2$, $p_2=3$ and $p_1=2$.
Since $1+p_i$ is divisible by $p_n$, we see that $1+p_i\geq p_n$ and $p_n\neq p_i$, which gives $p_n\geq1+p_i$.
Thus, $p_n=1+p_i,$ which is possible, when $p_i=2$ and $p_n=3$.