A positive integer $\ n\ $ is called totient , if there is a positive integer $\ m\ $ such that $\ \varphi(m)=n\ $ holds , where $\ \varphi(m)\ $ is the totient function.
Is $\ k!\ $ totient for every positive integer $\ k\ $?
For $\ 2\le k\le 200\ $ I could find positive integers $\ a,b\ $ with $\ a\cdot b=k!\ $ such that $\ a+1\ $ and $\ b+1\ $ are both (proven) primes. If we rely on the BPSW-test, I arrived at $\ k=500\ $.
Heuristically, we should be able to find $\ a,b\ $ in every case, but I think this cannot be proven. Is there another way to prove that every factorial is totient ?
Here the PARI/GP code searching a solution :
gp > for(n=2,20,s=n!;t=0;gef=0;while(gef==0,t=t+1;if(Mod(s,t)==0,if(isprime(t+1,2)==1,if(isprime(s/t+1,2)==1,z=(t+1)*(s/t+1);gef=1;print(n," ",z))))))
2 6
3 14
4 39
5 183
6 905
7 7563
8 60483
9 393133
10 4233607
11 79833602
12 526901771
13 9340531203
14 101708006407
15 1438441804811
16 31384184832003
17 414968666112007
18 6499379367936067
19 123488207990784067
20 2513998741782528031
gp >
Best Answer
Yes, look at https://artofproblemsolving.com/community/c6h140361. Basically, we can choose $$n=\frac{k!\cdot k\#}{\varphi( k\#) } ,$$ where $k\# = \prod_{p \leq k} p$ is the primorial. To see why this works, let $$ k!=p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r} $$ be the unique prime factorization of $k!$, then $k\#=p_1p_2\cdots p_r$ and $\varphi(k\#)=$$(p_1-1)(p_2-1)$$\cdots (p_r-1)$. Each of $p_i-1 \leq k$, and so $\varphi(k\#) \mid k!$. Furthermore, let $\varphi(k\#)=p_1^{l_{1}}p_2^{l_{2}}\cdot p_r^{l_{r }}$, then $$n=p_1^{e_1+1-l_1}p_2^{e_2+1-l_2}\cdots p_r^{e_r+1-l_r},$$ and so it's easy to verify $\varphi(n)=p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}=k!.$