The problem asks us to disprove that there exists an odd integer $x$ such that the following expression is divisible by 8:

$8 \nmid \left( x^2 + (x + 6)^2 + 6 \right)$

Step-by-Step Analysis:

1. Expression breakdown: The given expression is:

   $x^2 + (x + 6)^2 + 6$

   Expanding $(x + 6)^2$:

   $(x + 6)^2 = x^2 + 12x + 36$

2. Substitute into the original expression:

   $x^2 + (x + 6)^2 + 6 = x^2 + \left( x^2 + 12x + 36 \right) + 6$ Simplifying this:

   $= 2x^2 + 12x + 42$

3. We need to prove that $8 \nmid \left( 2x^2 + 12x + 42 \right)$ for any odd $x$.

Step 1: Express $x$ as an odd integer.

An odd integer can be written as $x = 2k + 1$, where $k$ is an integer.

Step 2: Substitute $x = 2k + 1$ into the expression:

$2x^2 + 12x + 42 = 2(2k + 1)^2 + 12(2k + 1) + 42$

Expanding $(2k + 1)^2$:

$(2k + 1)^2 = 4k^2 + 4k + 1$

Now, substitute this into the expression:

$2(4k^2 + 4k + 1) + 12(2k + 1) + 42 = 8k^2 + 8k + 2 + 24k + 12 + 42$

Simplifying:

$= 8k^2 + 32k + 56$

Step 3: Factor the expression:

$8k^2 + 32k + 56 = 8(k^2 + 4k + 7)$

Step 4: Check for divisibility by 8.

The expression is clearly divisible by 8 for any integer $k$. This implies that for all odd integers $x$, the expression is divisible by 8.

Conclusion:

We have shown that for any odd $x$, the expression $x^2 + (x + 6)^2 + 6$ is divisible by 8. Therefore, the statement that there exists an odd integer $x$ such that the expression is not divisible by 8 is false.

Thus, it is impossible to disprove that the expression is divisible by 8 for odd $x$, because it is always divisible by 8 for odd integers.

Would you like more details on any part of this explanation or have any questions? Here are some related questions for further exploration:

1. How would the solution change if $x$ were an even integer instead of odd?
2. What are the general properties of divisibility by 8 for quadratic expressions?
3. Can we generalize this problem to higher powers of integers and moduli other than 8?
4. What are some common techniques to disprove divisibility claims?
5. How do odd and even numbers affect divisibility in polynomial expressions?

Tip: When working with divisibility problems involving quadratic expressions, expressing integers in terms of their modular properties (odd/even, mod $n$) can simplify calculations significantly!