# An equation concerning perfect numbers

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 :(

• Where you took this problem? Feb 27, 2019 at 11:47
• Can you explain what are you asking for? Feb 27, 2019 at 12:38
• What is a source of this problem? Feb 27, 2019 at 13:14
• My creativity...(not joking)... Feb 28, 2019 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 $$p_n=3$$.

• Can you elaborate? Feb 27, 2019 at 12:43
• @MATHS MOD I added something. See now. Feb 27, 2019 at 13:12

Let $$p_1,\ldots,p_n$$ be distinct primes, with $$p_1, 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. $$\blacksquare$$