# An equation concerning perfect numbers

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 :(

• Where you took this problem? – Michael Rozenberg Feb 27 at 11:47
• Can you explain what are you asking for? – MATHS MOD Feb 27 at 12:38
• What is a source of this problem? – Michael Rozenberg Feb 27 at 13:14
• My creativity...(not joking)... – 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 $$p_n=3$$.