MHB How to Submit Solutions for POTW #330?

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Here is this week's POTW:

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For $n$ a positive integer, let $\phi(n)$ be the number of positive integers not exceeding $n$ and coprime to $n$. Show that every composite number $n \equiv 1\pmod{\phi(n)}$ has at least three distinct prime divisors. (In fact, $n$ would have at least four distinct prime divisors. You can prove this harder result if you like.)
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No one answered this week's problem. You can read my solution below.

Spoiler

By way of contradiction, suppose $n$ has only two distinct prime factors. Since $n \equiv 1 \pmod{\phi(n)}$, $n$ is not divisible by the square of any prime, and thus $n = p_1 p_2$ for some distinct primes $p_1, p_2$. Consider the integer
$$\frac{n-1}{\phi(n)} = \frac{p_1p_2 -1}{(p_1 - 1)(p_2 - 1)} = 1 + \frac{1}{p_1 - 1} + \frac{1}{p_2 - 1}$$ The latter expression is greater than $1$ and no greater than $2.5$, so being an integer
$$1 + \frac{1}{p_1 - 1} + \frac{1}{p_2 - 1} = 2$$
This forces $p_1 = p_2 = 3$, contradicting the assumption that $p_1 \neq p_2$.