If m is a positive integer, then `[(sqrt(3)+1)^(2m)]+1`, where [x] denotes greatest integer `lex`, must be divisible by

Options

A. `2^(m)`
B. `2^(m+1)`
C. `2^(m+2)`
D. `2^(2m)`

Explanation

To solve the problem, we need to analyze the expression \([( \sqrt{3} + 1)^{2m} ] + 1\), where \([x]\) denotes the greatest integer function.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \((\sqrt{3} + 1)^{2m}\). We know that \(\sqrt{3} + 1\) is greater than 1, and thus \((\sqrt{3} + 1)^{2m}\) will be a positive integer for any positive integer \(m\).

2. **Finding the Conjugate**:
We can also consider the conjugate \((\sqrt{3} - 1)^{2m}\). Since \(\sqrt{3} - 1\) is less than 1, \((\sqrt{3} - 1)^{2m}\) will be a small positive number that approaches 0 as \(m\) increases.

3. **Using the Binomial Theorem**:
We can expand \((\sqrt{3} + 1)^{2m}\) using the binomial theorem:
\[ (\sqrt{3} + 1)^{2m} = \sum_{k=0}^{2m} \binom{2m}{k} (\sqrt{3})^k (1)^{2m-k} \]
This expansion will yield integer terms.

4. **Finding the Integer Part**:
The integer part of \((\sqrt{3} + 1)^{2m}\) can be expressed as:
\[ [(\sqrt{3} + 1)^{2m}] = (\sqrt{3} + 1)^{2m} + (\sqrt{3} - 1)^{2m} - 1 \]
This is because the fractional part \((\sqrt{3} - 1)^{2m}\) is very small and contributes less than 1.

5. **Adding 1**:
Now, we add 1 to the integer part:
\[ [(\sqrt{3} + 1)^{2m}] + 1 = (\sqrt{3} + 1)^{2m} + (\sqrt{3} - 1)^{2m} \]

6. **Analyzing the Result**:
The expression simplifies to:
\[ (\sqrt{3} + 1)^{2m} + (\sqrt{3} - 1)^{2m} \]
This is an integer since both terms are integers.

7. **Divisibility**:
We need to check the divisibility of this expression. Notice that:
\[ (\sqrt{3} + 1)^{2m} + (\sqrt{3} - 1)^{2m} \]
can be shown to be divisible by 2 for any positive integer \(m\).

### Conclusion:
Thus, we conclude that \([( \sqrt{3} + 1)^{2m} ] + 1\) must be divisible by 2.