Timeline for Is the number of solutions of $\phi(x)=n!$ bounded? If yes, what is its bound?
Current License: CC BY-SA 4.0
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| Jul 9, 2020 at 18:26 | history | edited | Gerhard Paseman | CC BY-SA 4.0 |
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| Jul 8, 2020 at 5:05 | comment | added | Gerhard Paseman | Toward a lower bound, which is the main thrust of the question. Note first that for P the product of the primes at most n one has $\phi((n!)P/\phi(P))=n!$ and one has an alternate representation when n+1 is an odd prime. There also is a representation involving a prime q bigger than n whenever $\phi(q)$ divides $n!/(P\phi(P))$. Proving the existence of $q$ for every sufficiently large n should show the number of solutions grows without bound as $n$ grows. Gerhard "Is Almost Convinced Of Unboundedness" Paseman, 2020.07.07. | |
| Jul 8, 2020 at 3:39 | comment | added | Gerhard Paseman | If we let k be the exponent of the exact power of two dividing m, we get that any solution to the general equation must (for k not too small) be less than 3(k+1)m, giving a weak but non exponential (in m) upper bound on the number of solutions. Gerhard "Really Likes This Re-engineering Stuff" Paseman, 2020.07.07. | |
| Jul 8, 2020 at 3:30 | history | edited | Gerhard Paseman | CC BY-SA 4.0 |
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| Jul 8, 2020 at 3:17 | history | answered | Gerhard Paseman | CC BY-SA 4.0 |