How Can You Find All Integer Pairs of (m, n) Given the Equation (1 cdot m cdot m^2 cdot m^3 p^n)?

Let's explore how to find all integer pairs of (m, n) given the equation (1 cdot m cdot m^2 cdot m^3 p^n), where (p) is a prime number. We will break down the problem into several cases, each addressing a specific scenario.

1. Case: (n) is a Positive Integer

Given the equation (1 cdot m cdot m^2 cdot m^3 p^n), we can rewrite it as (m^6 p^n). This implies that (m^6) must be an integer power of a prime number (p). Since (m^6 (p)^{n/6}), (m) must be an integer power of (p), so let (m p^a) for some integer (a).

Simplification of the Equation:

Substitute (m p^a) into the original equation:

(1 cdot p^a cdot (p^a)^2 cdot (p^a)^3 p^n)

(1 cdot p^a cdot p^{2a} cdot p^{3a} p^n)

(p^{6a} p^n)

This implies (6a n), so (n 6a).

Now, consider the case where (p 2): (1 cdot 2^a cdot (2^a)^2 cdot (2^a)^3 2^n) (1 cdot 2^a cdot 2^{2a} cdot 2^{3a} 2^n) (2^{6a} 2^n) (n 6a) This gives us integer solutions such as ( (0, 0), (1, 6), (2, 12), ldots ).

2. Case: (n 0)

If (n 0), the equation becomes:

(1 cdot m cdot m^2 cdot m^3 p^0)

(1 cdot m cdot m^2 cdot m^3 1)

(m^6 1)

(m pm 1)

For (m 1), we get ( (1, 0) ), and for (m -1), we get ( (-1, 0) ). These are trivial integer pairs.

3. Case: (n) is a Negative Integer

If (n) is a negative integer, the right-hand side of the equation becomes a positive fraction less than 1, whereas the left-hand side (1 cdot m cdot m^2 cdot m^3 m^6) is an integer for all integer values of (m). Thus, the equation is not valid in this case.

Summary of Solutions:

The only integer pairs for which the equation (1 cdot m cdot m^2 cdot m^3 p^n) is valid are:

In conclusion, the integer pairs ((m, n)) satisfying the equation (1 cdot m cdot m^2 cdot m^3 p^n) depend on the value of (p) and are generally limited to ((0, 0)) for any prime (p) and additional pairs for (p 2).