Find all n such that $\varphi(n) = n/24$
This is my attempt:
$24 = 2^3*3$
$\varphi(24^k) = (2^{3k}-2^{3k-1})(3^k-3^{k-1})=2^{3k}(1-1/2)*3^k(1-1/3)=2^{3k}*3^{k-1}$
$24^k/24 = 24^{k-1}= 2^{3k-3}*3^{k-1}$, For $k= 1, 2, 3, ...$
I also tried multiplying $24^k$ by a factor but I didn't get anywhere. Is there a formula to find such n?
Any help would be appreciated!
If $n$ is written as a product of primes, $n=p^\alpha q^\beta\cdots$ with $\alpha,\beta\cdots>0$, then $n=24\phi(n)$ becomes $$p^\alpha q^\beta\cdots=24p^{\alpha-1}(p-1)q^{\beta-1}(q-1)\cdots$$ and so $$pq\cdots=24(p-1)(q-1)\cdots\ .$$ Now the LHS is a product of distinct primes, but the RHS has a factor of $2$ twice. (In fact, three times... in fact, possibly more.) So there is no solution.
