Assume that there are infinitely many Mersenne primes, prove there are infinitely many positive integers m and n such that $\phi(m)=\sigma(n)$.

Question

Conjecture 3 of chapter 1 states that there are infinitely many Mersenne prime numbers. Mersenne prime numbers are integers that are in the form of $2^p-1$ where p is prime.The first 3 Mersenne prime numbers are $2^2-1=3$, $2^3-1=7$, $2^5-1=31$.Let p be prime then $\phi(2^p-1)=2^p-1-1=2^p-2$ and $\sigma(2^p-1)=2^p-1+1=2^p$

I feel like I'm on the right lines but I just can't put the last bit together. Could someone give me some ideas, please,

($\phi(m)$ is the number of integers relatively prime to m and $\sigma(n)$ is the sum of the divisors of n.

Answer 1

Hint: If you're supposed to use the existence of infinitely many Mersenne primes, it's likely that the construction of either $m$ or $n$ will involve Mersenne primes, but not necessarily both. Can you find a number with $\sigma(n)=2^p-2$? How about a number with $\phi(m)=2^p$? It may be useful to remember how $\phi$ and $\sigma$ relate to prime factorizations, to find a number $m$ or $n$ with a nice prime factorization and the desired property.