Let n>1 be an integer. Prove that φ (n) | n if and only if n is of the form 2^a 3^b where a≥1 and b≥0 are integers.
$\displaystyle \varphi(2^a3^b)=2^a3^{b-1} | 2^a3^b$
$\displaystyle \frac{n}{\varphi(n)} = \prod_{i=1}^{i=r}\frac{p_i}{p_i-1}$
the rhs denominator is dividable by $\displaystyle 2^r$ if n is not dividable by 2 or $\displaystyle 2^{r-1}$ if it is dividable by 2
the rhs numerator is dividable only by 2 if n is dividable by a power of 2
hence for the lhs to become an integer and thus $\displaystyle \varphi(n)|n$ the number of primes in the factorization of n can be at most 2
hence
$\displaystyle \frac{pq}{(p-1)(q-1)} $ must be integer
therefore $\displaystyle \frac{pq}{(p-1)(q-1)} \ge 2$ equivalent to
(*) $\displaystyle 2 \ge (p-2)(q-2)$
wlog q<p
examine the two cases:
1) q=2: $\displaystyle \frac{pq}{(p-1)(q-1)} = \frac{2p}{p-1}$ is integer only for $\displaystyle p=3$
2) $\displaystyle q \ge 3$: then (*) becomes $\displaystyle 2 \ge (p-2)(q-2) > p-2$ thus $\displaystyle p \le 4$ since p prime p = 2 or 3 wich is a contradiction to q<p