As in ...
Euler's totient function - Wikipedia, the free encyclopedia
... if...
(1)
... where the are distinct primes, then...
(2)
Merry Christmas from Italy
The problem is:
for i=1...k
where , are odd primes in increasing order, and are positive integers.
We're told that for some odd, composite n.
Express the value of n in terms of and .
I know since n is composite is not equal to (n-1), but I'm not sure what to do elsewhere.
As in ...
Euler's totient function - Wikipedia, the free encyclopedia
... if...
(1)
... where the are distinct primes, then...
(2)
Merry Christmas from Italy
I apologize because I didn't undestand exactly Your question ...
For the fundamental theorem of arithmetic any m can be expressed as...
(1)
... so that any m doesn't have 'special properties'. If I understand correctly Your problem is, given m, to find the value of n for which is...
(2)
In general the equation (2), that supplies the 'inverse' of , doesn't have a single solution and can be solved by 'systematic search'...
Merry Christmas from Italy
In the table reported in ...
Euler's totient function - Wikipedia, the free encyclopedia
... You can verify that is an even number and the only n for which doesn't devide are and , being p a prime number grater than 2. That is consequence of the basic property that if m and n are coprime, then ...
Merry Christmas from Italy
Ah yes, I see that now, thank you. However, in the problem it is stated that n is odd and composite. This means that neither of those solutions are valid for n, since p is prime and 2*p is even.
I believe I figured it out: n must only have 1 odd factor otherwise 2^k, where k is the number of odd factors of n, would have to divide m. Since m only has one factor of 2, then k=1. Since phi(n)=m, that odd prime must be 3, otherwise an additional factor of 2 would appear from phi(q^c)=q^c*(1-1/q), where n = q^c.
Therefore n = p^(e+1) = 3^(e+1), where p and e are the prime factor and power of m.
Thanks!