An equation concerning perfect numbers

Question

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 :(

Answer 1

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 $p_n=3$.

Answer 2

Let $p_1,\ldots,p_n$ be distinct primes, with $p_1<p_2<p_3<\ldots<p_n$, satisfying

$$ (1+p_1)(1+p_2)(1+p_3) \cdots (1+p_n) = 2p_1p_2p_3 \cdots p_n. $$

If $p_1>2$, the RHS is of the form $2m$, with $m$ odd. Since each factor $1+p_i$ is even, there can only be one factor. But then $1+p_1=2p_1$, which is impossible.

Thus, $p_1=2$, and since $3$ divides the RHS, $p_2=3$. Since each factor $1+p_i$ is even for $i>1$, and $1+p_2=4$, the LHS is a multiple of $8$ whereas the RHS is not a multiple of $8$ if $n>2$. So $n=2$, and we have $p_1=2$, $p_2=3$ as the only solution. $\blacksquare$