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Using the product formula (the formula with the prime factors of $n$), I got $$1=6\frac{(P_1-1)}{P_1}\frac{(P_2-1)}{P_2}\cdots\frac{(P_k-1)}{P_k}\,.$$

Batominovski
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Megan
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    Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Additionally, if you are looking for prsomebody to check your proof, please update your post to include the tag [tag:proof-verification]. – platty Dec 12 '18 at 00:19

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i don't think the proof is enough, as it stands. There is a small ingredient. We are taking the product, with one or more primes, of $\frac{p-1}{p}.$

There are two types of potential problems. Some cancellation is possible, putting the product in lowest terms may be a bit surprising. Second, we can get the product arbitrarily close to $\frac{1}{6}$ byt choosing many large primes.

However: the collection of chosen primes is finite, and one of them is largest, call that $q.$ For any of the other, smaller primes, since $p<q,$ we also have $p-1 < q.$ That means that the numerator of the product fraction is not divisible by $q.$ When we put the fraction in lowest terms, the denominator is still divisible by $q.$

For your problem, this means that the largest prime involved, $q,$ must be a factor of $6.$ The possibilities for the product are just: $$ \{ \}, \frac{1}{1} $$ $$ \{ 2\}, \frac{1}{2} $$ $$ \{ 3\}, \frac{2}{3} $$ $$ \{ 2,3\}, \frac{1}{3} $$

Will Jagy
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Hints:

We know that if $\;p_1,...,p_k\;$ are the different primes diving $\;n\;$ , then

$$\varphi(n)=n\prod_{j=1}^k\left(1-\frac1{p_j}\right)$$

Well, take it now...And I couldn't completely understand what you wrote, but I think you're on the right track: there is no solution.

DonAntonio
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  • i used the same definition but because i couldnt type the product symbol i used the definition of the product....anyway thanks for verifying that there is no solution – Megan Dec 12 '18 at 00:27