Find all positive primes $p_1,p_2,p_3, \cdots p_n$ such that $$(1+\frac{1}{p_1})(1+\frac{1}{p_2}) \cdots (1+\frac{1}{p_n}) =2$$

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 required 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 :(

  • $\begingroup$ Where you took this problem? $\endgroup$ – Michael Rozenberg Feb 27 at 11:47
  • $\begingroup$ Can you explain what are you asking for? $\endgroup$ – MATHS MOD Feb 27 at 12:38
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    $\begingroup$ What is a source of this problem? $\endgroup$ – Michael Rozenberg Feb 27 at 13:14
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    $\begingroup$ My creativity...(not joking)... $\endgroup$ – MATHS MOD Feb 28 at 2:16

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 $p_n=3$.

  • $\begingroup$ Can you elaborate? $\endgroup$ – MATHS MOD Feb 27 at 12:43
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    $\begingroup$ @MATHS MOD I added something. See now. $\endgroup$ – Michael Rozenberg Feb 27 at 13:12

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