Solving Equations Involving Euler's Totient Function: φ(n) k

Equations of the form φ(n) k, where φ is Euler's Totient Function, can be approached using a variety of methods. This function, φ(n), counts the number of positive integers up to n that are coprime to n. Below, we will explore a general approach to solving such equations, including the definition, properties, and application of these methods.

Understanding Euler's Totient Function

Euler's Totient Function, denoted as φ(n), is defined for a positive integer n with the prime factorization n p_1^{k_1}p_2^{k_2}cdots p_m^{k_m}. The formula for φ(n) is given by:

φ(n) nleft(1 - frac{1}{p_1}right)left(1 - frac{1}{p_2}right)cdots left(1 - frac{1}{p_m}right)

Properties of Euler's Totient Function

Multiplicative Property: If a and b are coprime, then φ(ab) φ(a)φ(b). Prime Power Property: If n is a prime p, then φ(p) p - 1. Power of Prime Property: If n p^k, then φ(p^k) p^k - p^{k-1} p^{k-1}(p - 1).

General Steps to Solve φ(n) k

The general approach to solving equations of the form φ(n) k involves several steps, as outlined below:

Factor k

Start by factoring k into its prime factors. This can provide insights into the possible values for n.

Check Possible Forms of n

Single Prime: If n is a prime p, then φ(p) p - 1. Thus, if k - 1 is a prime number, then n k 1. Power of a Prime: If n p^k, then set φ(p^k) k and solve for p and k. Product of Distinct Primes: For n p_1^{k_1}p_2^{k_2}cdots p_m^{k_m}, calculate φ(n) using the formula and compare to k.

Trial and Error

In cases where k is small, it may be feasible to try small values of n to see if φ(n) k.

Use Known Values

Consult tables and lists of known values for φ(n) for small integers. These can provide quick references to see if a solution exists.

Consider the Range of n

The function φ(n) is non-increasing for n as a product of more primes. Therefore, if k is large, n must also be large, and vice versa.

Example: Solving φ(n) 8

To solve the equation φ(n) 8, follow these steps:

Single Prime: n p is not possible since φ(p) p - 1 gives no prime p such that p - 1 8. Power of a Prime: For p 3, φ(3^2) 3^2 - 3 6 is too low for p 5, φ(5^2) 5^2 - 5 20 is too high. Product of Distinct Primes: For n 15, φ(15) 15left(1 - frac{1}{3}right)left(1 - frac{1}{5}right) 15 cdot frac{2}{3} cdot frac{4}{5} 8.

Thus, n 15 is a solution.

Conclusion

There is no single formula for solving φ(n) k. However, by understanding the properties of the totient function and systematically checking possible forms of n, you can find solutions. For larger values of k or specific cases, computational methods may also be useful.