Solving the Equation: 100a10bc^2 A^2 abc^5 B^5

In this article, we will explore an intricate problem involving number theory, specifically focusing on integer solutions and perfect squares. The equation we are tasked with solving is:

1. The Initial Equation

Given the equation:

100a 10b c^2 a b c^5

Our goal is to find the integer values of (a), (b), and (c) that satisfy this equation. Let's break down the problem step by step.

2. Simplifying the Problem

We can simplify this problem further by defining:

(n 100a 10b c)

(s a b c)

Substituting these into our original equation, we get:

n^2 s^5

This implies that (n) is a perfect square and (s) is a positive integer such that (s^5) is a perfect square. Consequently, (s) itself must be a perfect square.

3. Finding Possible Values of (s)

To find the maximum possible value of (a b c), we note that:

a b c 10 10 10 30

Since (s) must be a perfect square and must be a positive integer less than 10, the possible values for (s) are 1, 4, and 9.

3.1. Checking (s 1)

s^5 1^5 1 quad implies quad n^2 1 quad implies quad n 1

This is not possible because (a b c) must be positive integers.

3.2. Checking (s 4)

s^5 4^5 1024 quad implies quad n^2 1024 quad implies quad n 32

This is not possible because (n leq 1000) when (a b c leq 10).

3.3. Checking (s 9)

s^5 9^5 59049 quad implies quad n^2 59049 quad implies quad n 243

This is also not possible because (n leq 1000).

4. Further Analysis

We need to consider other possible values of (s) that are less than 10. Let's check the cases for (s 3, 5, 6, 7, 8) and find that none of these values satisfy the equation.

5. Final Solution

The only feasible solution is when (s 9), which leads us to:

s 9 quad and quad n 243

Then we need to find (a b c leq 10) such that:

100a 10b c 243

This gives:

a 2 quad b 4 quad c 3

Thus, we calculate:

ab - c 24 - 3 5

Therefore, the final answer is:

boxed{5}

Additional Insights

Let's re-analyze the problem using the concept of number theory and prime factorization. If we let:

A prod{p_i^{alpha_i}} quad and quad B prod{p_i^{beta_i}}

The equation becomes:

A^2 B^5 quad Rightarrow quad 2alpha_i 5beta_i

This implies that 5 must divide (alpha_i) and 2 must divide (beta_i). Thus:

alpha_i 5alpha_i quad and quad beta_i 2beta_i

Thus, (alpha_i beta_i), and we can let:

D prod{p_i^{alpha_i}}

This gives:

A D^5 quad and quad B D^2

Further analysis shows:

A 100abc leq 1000 Rightarrow D leq 3

Considering (D 1), (D 2), and (D 3), only (D 3) gives a viable solution:

A 3^5 243 quad Rightarrow quad a 2 quad b 4 quad c 3

Thus, the only solutions are:

001 quad and quad 243