1) find all positive integers solutions of phi(n)=12

Question

eva12 · Answer

I know the divisors of 12 :1,2,3,4,6,12.  Possible factors :2,3,4,5,7,13  how do I find possible solutions?

vats-sagar · Answer

phi(n)     ?

anonymous · Answer

its the totient function right?

eva12 · Answer

Yes

anonymous · Answer

one solution will be 13

eva12 · Answer

Is that since 13-1 =12 => 12*1 since phi (13)* phi(2) =12*1 right?

anonymous · Answer

yes for a prime number say p$$\phi(p)=p-1$$and 13 satisfies the case but there can be more solutions

anonymous · Answer

when n was prime the solution we got was 13 :)
prime factorization of n if (n is not prime of course)->$$n=p_1^{x_1} \times p_2^{x_2} \times p_3 ^{x_3}......$$where $p_1,~p_2,...$ are primes and $x_1,~x_2,.... \ge 1$
now we know that totient function is multiplicative so ->$$\phi(n)=\phi(p_1^{x_1}) \times \phi(p_2^{x_2}).....$$since $p_1,~p_2,... ~are~primes~we~can~write~it~as->$$$\phi (n)=\left( p_1^{x_1}-p_1^{x_1-1} \right)\left(p_2^{x_2}-p_2^{x_2-1}  \right)......$$
$$12=\left( p_1^{x_1}-p_1^{x_1-1} \right)\left(p_2^{x_2}-p_2^{x_2-1}  \right)......$$
we know that 
12=1x12
    =2x6
    =3x4
    
so we can have only 3 cases when n is not prime
now you just have to evaluate these cases and then u can get more solutions from here

loser66 · Answer

It is a very long solution for this. ha!! But let me try
$n =\prod_{i=1}^s p_i^{a_i}$, $\phi (n) = \prod_{i=1}^s p_i^{a_i -1} (p_i-1)$

Hence $(p_i -1)| \phi (n)=12$ . Therefore, if $p_i > 17, p_i-1 =16 > 12\rightarrow 16\cancel | 12$

Hence $p_i \leq 13$
then $\large n = 2^a~ 3^b~ 5^c~ 7^d ~11^e ~13^f$
 
Moreover, $p_i^{a_i} | n$  , hence 
 
1) $2^{a-1} |12 \implies (a-1)\leq 2, hence~~ a =0,1,2,3 $
2) $3^{b-1}| 12\implies b-1\leq1,~hence~ b=0,1,2$
3) $5^{c-1}|12\implies c-1 \leq 0~hence c =0,1 $  
4) $7^{d-1}|12\implies d-1\leq 0~hence~d=0,1$
5) $11^{e-1}|12\implies e-1\leq 0~hence~e =0,1$
6)$13^{f-1}|12\implies f-1\leq 0, ~~hence~f =0,1$
ha!!! now combine all
If a =b=c=d=e= 0 $f\neq 0$ , that is f =1, hence n = 13, $\phi(13) =12\color{red}{\checkmark}$
If a=b=c =d =f =0, $e\neq 0$ , that is e =1, hence n = 11, but $\phi (11)\neq 12$ reject!!
If a=b=c=e=f =0 $d \neq 0$........
You do all of the options, the final result are n = 13,21,26,28, 36,42
Good luck.

eva12 · Answer

Thank you