How to Find Integer Solutions for the Equation 2^{2016} 2^{2012} 2^{2008} 2^m n^2

Let's explore how to find the integer solutions for the equation (2^{2016} cdot 2^{2012} cdot 2^{2008} cdot 2^m n^2). This problem involves analyzing exponents and square properties. We'll break down the solution step-by-step.

Initial Simplification

First, let's simplify the left-hand side of the equation:

Hence we have:

[2^{2016} cdot 2^{2012} cdot 2^{2008} 2^{2008} cdot 12^4 cdot 2^8 2^{2008} cdot 273]

Here, (2^{2008}) is already a square. For the rest, we need to determine if adding or adjusting (2^{273}) can make it a square. It turns out that adding (16 cdot 2^{273} 2^{283}) to (2^{273}) makes it a perfect square. Therefore:

[2^m 2^{283} 2^{2012}]

Therefore, we find that:

[m 2012]

Algebraic Rearrangement and Analysis

We can further rearrange the original equation:

[2^{2016} cdot 2^{2012} cdot 2^{2008} cdot 2^m 2^{2008} cdot 273 cdot 2^m n^2]

We assume (m 2008x), giving us:

[2^{2008} cdot 273 cdot 2^{2008x} n^2]

If (x 4), then:

[2^{2016} cdot 2^{2012} cdot 2^{2008} cdot 2^m 2^{1004} cdot 17^2 cdot 2^{2012}]

This provides a solution where (m 2012) and (n 2^{1004} cdot 17). Another solution, considering (n geq 0), would be:

[n 2^{1004} cdot 23]

Other solutions are more complex and require deeper analysis, but we'll focus on the simpler cases.

Proof for Specific Cases

Next, let's prove that there are no solutions when (m

Consider the left-hand side of the equation:

[2^{2016} cdot 2^{2012} cdot 2^{2008} cdot 2^m 2^m cdot 2^{2008} cdot 12^4 cdot 2^8]

We factor (2^{2008}) and (12^{2008-m} cdot 2^{2012-m} cdot 2^{2016-m}) where the latter part is odd. Therefore, (2^m) and the odd part must be squares, implying that (m) is even and the odd part is (n_1^2).

Furthermore, (n_1^2 - 1 2^l cdot 273), where (l 2008 - m) is even. This gives:

[n_1 - 1 cdot n_1 1 2^l cdot 273]

We then analyze two sub-cases:

Sub-case 1: (n_1 - 1 2^{l-1} cdot d), (n_1 1 2 cdot e)

Here, (273 de), and by checking (d), (e), we find no valid pair. Hence, this sub-case is eliminated.

Sub-case 2: (n_1 - 1 2 cdot d), (n_1 1 2^{l-1} cdot e)

This leads to more restrictive conditions, ultimately proving no valid solutions exist.

Thus, the only valid solutions are:

[m 2008, 2012, 2016] [n 2^{1004} cdot 17, 2^{1004} cdot 23]

This completes our exploration into finding integer solutions for the given equation.