For any integer n show that there exist integers k and j such that φ(k)+σ(j)=n
So, I would note that so taking , .
However, this is only for the natural numbers and zero...not the whole integers. You can't get a negative number of divisors!
It is also a somewhat boring result - why the Euler totient function if we can just forget about it?...
Have you perhaps missed out something from the question?
If p is prime, then sigma(p)=p+1 (since there are only the two factors, 1 and p). Also, phi(p)=p-1 since all of the integers from 1 to p-1 are coprime to p, so sigma(p)+phi(p)=2p.
I think it's probably true, but I can't see a way to prove that there are no composite n with sigma(n)+phi(n)=2n.
Take with and prime. ,Originally Posted by hollywood
Oops, I just wiped out all semiprimes
I believe there is a general proof that can be built around this with infinitely many prime factors instead of a semiprime. I've seen it but unfortunately forgot it, I'll have to check it out some time.
Conclusion : only prime numbers satisfy this statement, along with 1, assuming the validity and existence of the GFP (Generalized Forgotten Proof )