Positive Integer Solutions to m - n^2 4m^2n

In this article, we explore the positive integer solutions to the equation (m - n^2 4m^2n). Through a detailed analysis and the application of number theory, we derive the solution set and provide a thorough explanation of the steps involved.

Introduction

This article delves into the positive integer solutions of the given equation. The approach involves examining the properties of primes and the highest powers of primes dividing integers. The discussion will be structured to guide the reader through each logical step.

The Equation and Prime Division

Consider the equation (m - n^2 4m^2n). Let (e_{p, n}) denote the highest power of a prime (p) dividing (n). We state that (p^{e_{p, n}} mid n) and (p^{e_{p, n} 1} mid n).

Suppose (m) and (n) are positive integers such that (m - n^2 4m^2n). Fix a prime (p) and let (e_{p, m} alpha) and (e_{p, n} beta).

Case Analysis: (beta alpha)

Assume (beta alpha). Then, (m p^alpha a) and (n p^beta b) with (p mid ab). This implies (m - n p^alpha big(a - p^{beta - alpha} bbig)) and (n^2 - m p^alpha big(p^{2beta - alpha} b^2 - abbig)) with (p) dividing neither (a - p^{beta - alpha} b) nor (p^{2beta - alpha} b^2 - a) due to (beta alpha) and (p mid ab).

The highest power of (p) dividing (m - n^2) and (n^2 - m) leads to the conclusion that (beta leq alpha). Therefore, (e_{p, n} leq e_{p, m}) for each prime (p), implying that (n mid m).

Reformulating the Equation

Assume (m kn) with (k in mathbb{N}). Substitute into the original equation and simplify to get:

(n k frac{4k^2}{k^2 - 1}).

The second term on the right-hand side must be an integer, hence (k^2 - 1 mid 4k^2). Given that (gcd(k^2 - 1, k) 1), we must have (k^2 - 1 mid 4). Thus, (k 2) or (k 3).

For (k 2) and (k 3), we get (n 18) and (m 36) or (n 12) and (m 36). The only solutions in positive integers are: (m 36) and (n 12) or (n 18).

No Prime Divisor Differences

There can be no prime that divides (m) but not (n) nor vice versa. If (p mid m) and (p mid n), then (p mid m - n) and (p mid n^2 - m). This contradicts the requirement that any prime dividing (m) or (n) must divide at least one of the right-hand terms.

Proving (m eq n)

We can have (m eq n). If (m nq) with (q geq 1), then, using the original equation and simplifying, we derive (n q frac{4q^2}{q^2 - 1}). The second term on the right-hand side must be an integer, hence ((q^2 - 1) mid (4q^2)). Given (gcd(q^2 - 1, q) 1), we must have ((q^2 - 1) mid 4). Thus, (q 2) or (q 3).

For (q 2) and (q 3), we get (n 12) and (m 36) or (n 18) and (m 36). The only solutions in positive integers are: (m 36) and (n 12) or (n 18).

In conclusion, the only positive integer solutions to (m - n^2 4m^2n) are: (m 36) and (n 12) or (n 18).