prove that if n is divisible by 2, then there exist integers x and y such that n = 6x + 4y

To prove that if \( n \) is divisible by \( 2 \), then there exist integers \( x \) and \( y \) such that \( n = 6x + 4y \), we start by noting the properties of even numbers and considering the linear combination \( 6x + 4y \).

### Step-by-Step Proof

1. **Assume \( n \) is divisible by 2**:
   - Since \( n \) is divisible by \( 2 \), it means that \( n \) is an even number.
   - Therefore, we can write \( n = 2k \), where \( k \) is an integer.

2. **Consider the form \( n = 6x + 4y \)**:
   - We need to show that there exist integers \( x \) and \( y \) such that \( n = 6x + 4y \).
   - The expression \( 6x + 4y \) can be factored as:
     \[
     6x + 4y = 2(3x + 2y)
     \]
   - This shows that \( 6x + 4y \) is always divisible by \( 2 \), regardless of the values of \( x \) and \( y \).

3. **Expressing \( n = 2k \) in the form \( 6x + 4y \)**:
   - Since \( n = 2k \), we need to find values of \( x \) and \( y \) such that:
     \[
     2k = 6x + 4y = 2(3x + 2y)
     \]
   - Dividing both sides by \( 2 \), we get:
     \[
     k = 3x + 2y
     \]
   - This is a linear Diophantine equation, which has integer solutions for \( x \) and \( y \) for any integer \( k \). This means that for any even \( n = 2k \), we can always find integers \( x \) and \( y \) that satisfy the equation.

4. **Conclusion**:
   - Since \( k = 3x + 2y \) has integer solutions for \( x \) and \( y \), it follows that for any even \( n \), there exist integers \( x \) and \( y \) such that \( n = 6x + 4y \).
   - Therefore, if \( n \) is divisible by \( 2 \), then \( n \) can be expressed in the form \( 6x + 4y \) for some integers \( x \) and \( y \).

### Summary
We have shown that for any even number \( n \), it can be represented as \( 6x + 4y \), confirming the claim. The key idea was to express \( n \) as \( 2k \) and show that the equation \( k = 3x + 2y \) always has integer solutions for \( x \) and \( y \).

Would you like more details on the solution, or do you have any questions?

#### Related Questions:
1. How can we solve linear Diophantine equations generally?
2. What if \( n \) is not divisible by 2—can it still be represented by a similar equation involving \( 6x + 4y \)?
3. Can you provide an example where \( n \) is divisible by \( 2 \) and find corresponding values for \( x \) and \( y \)?
4. How does the proof change if \( n \) must also be divisible by another number, such as \( 3 \)?
5. How can we generalize this form to represent other even numbers using similar linear combinations?

#### Tip:
For any equation involving multiple unknowns, consider factoring or reducing the equation to see if a general pattern or common factor emerges. This can often simplify finding solutions.

Prove that if n is divisible by 2, then there exist integers x and y such that n = 6x + 4y.

Explore how any even number n can be expressed in the form 6x + 4y. This detailed proof breaks down key concepts from number theory and Diophantine equations, showing how integer solutions for x and y can always be found when n is divisible by 2. Suitable for advanced high school or undergraduate students.

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