Let σ(n) be the sum of the positive divisors of an integer n and ϕ(n) be the number of positive integers smaller than n that are coprime to n. If p is a prime number, what is the maximum value σ(p)ϕ(p)?
The actual value can be found at
http://www.proofwiki.org/wiki/Product_of_Sigma_and_Euler_Phi_Functions
Since the value is n^2 times a bunch of factors each less than 1, I guess n^2 is the maximum possible value.
To find the maximum value of σ(p)ϕ(p), we need to find the maximum possible values of σ(p) and ϕ(p) separately, and then multiply them together.
First, let's consider σ(p), which is the sum of the positive divisors of p. In the case of a prime number, p, the only positive divisors are 1 and p itself. Therefore, σ(p) = 1 + p.
Next, let's consider ϕ(p), which is the number of positive integers smaller than p that are coprime to p. Since p is a prime number, all positive integers smaller than p are coprime to p, except for the multiples of p itself. Therefore, ϕ(p) = p - 1.
Now, we can calculate the maximum value of σ(p)ϕ(p) by substituting the maximum possible values of σ(p) and ϕ(p) into the equation:
Maximum value = (1 + p) * (p - 1)
Simplifying this expression further, we get:
Maximum value = p^2 - 1
Therefore, the maximum value of σ(p)ϕ(p) is p^2 - 1 for a prime number p.