Solving IMO 2006 P4: A Comprehensive Approach to Integer Solutions

The International Mathematical Olympiad (IMO) problems are renowned for their complexity and the depth of mathematical insight required to solve them. In this article, we delve into the solution of IMO 2006 Problem 4, which involves finding all pairs of integers x, y such that 1 2^x 2^{2x-1} y^2.

Problem Statement

The problem statement is as follows: Determine all pairs of integers x, y such that 1 2^x 2^{2x-1} y^2.

Approach and Solution

Initial Observations

First, note that the problem asks for pairs of integers (not just positive integers). We can start by assuming that y is a positive integer.

Consider the case when x is negative. If x is less than -1 (i.e., -2 or less), the left side of the equation is not an integer. Therefore, we only need to consider x starting from -1 and onwards.

Small Values of x

Let's start by examining small values of x manually.

x -1: The left side is 2, which is not a perfect square. Therefore, no solutions.

x 0: The left side is 1 1 1 3, which is not a perfect square. Therefore, no solutions.

x 1: The left side is 1 2 4 7, which is not a perfect square. Therefore, no solutions.

x 2: The left side is 1 4 16 21, which is not a perfect square. Therefore, no solutions.

x 3: The left side is 1 8 64 73, which is not a perfect square. Therefore, no solutions.

x 4: The left side is 1 16 256 273, which is not a perfect square. Therefore, no solutions.

We continue this process and find:

x -1: The left side is 2, no solutions.

x 0: The left side is 4, which is a perfect square. Solutions: 0, 2 and 0, -2.

x 1: The left side is 7, no solutions.

x 2: The left side is 21, no solutions.

x 3: The left side is 73, no solutions.

x 4: The left side is 273, no solutions.

Algebraic Approach

We can simplify the equation by substituting a 2^x, leading to 1 a 2a^2 y^2. Since the left side is always odd (except for x 0, -1), we can set y 2b - 1. Inserting this substitution, we get:

1 a 2a^2 4b^2 - 4b.

Consider this as a quadratic in a. For it to have integer solutions, the discriminant 1 - 32b^2 b must be a perfect square. This leads to:

32b^2 - b k^2 - 1k - 1 for some integer k.

This equation is complex, and it doesn't use the fact that a is a power of 2. Hence, we consider another approach.

Factoring and Divisibility

Notice that 2^(x-1) divides both sides of the equation. This means that one of y-1 or y 1 must be divisible by 2^(x-1). Since y can't be much larger than 2^x, we can narrow down y to be one of 2^(x-1), 2^(x-1) 1, 2^(x-1) - 1.

Examining y 2^(x-1) - 1, we find that:

1 2^x 2^(2x-1) 2^(2x-2) - 3 cdot 2^(x-1) 1.

Setting this equal to y^2 and solving, we find:

x 4, leading to solutions x, y 4, 23 and x, y 4, -23.

Finally, we test the other possibilities and find that y 3 cdot 2^(x-1) 1 leads to a negative left side, which cannot be the case.

Thus, the complete set of solutions is:

(0, 2), (0, -2), (4, 23), (4, -23)

Conclusion

Through a combination of algebraic manipulation and divisibility arguments, we have successfully determined all integer solutions to the given problem. This problem exemplifies the importance of careful examination of each step and considering multiple approaches to reach a comprehensive solution.