The problem asks us to determine which of the given expressions must always be even, given that 'a' and 'b' are integers and their difference, $\text{a – b}$, is even.

To solve this, we first need to understand the implications of $\text{a – b}$ being an even number. If the difference between two integers is even, it means that both integers must have the same parity. There are two possible scenarios:

• Scenario 1: Both 'a' and 'b' are even integers.
• Scenario 2: Both 'a' and 'b' are odd integers.

We will examine each option under these two scenarios to see which expression consistently results in an even number.

Integers Parity Analysis

Let's recall some basic rules about even and odd numbers:

• $\text{Even} \pm \text{Even} = \text{Even}$
• $\text{Odd} \pm \text{Odd} = \text{Even}$
• $\text{Even} \pm \text{Odd} = \text{Odd}$
• $\text{Even} \times \text{Any integer} = \text{Even}$
• $\text{Odd} \times \text{Odd} = \text{Odd}$
• $\text{Even}^{\text{power}} = \text{Even}$ (for positive integer power)
• $\text{Odd}^{\text{power}} = \text{Odd}$ (for positive integer power)

Expressions Evaluation

Let's evaluate each option based on the two scenarios derived from $\text{a – b}$ being even.

Option 1: $\text{ab}$

• Scenario 1 (a is Even, b is Even): $\text{a} \times \text{b} = \text{Even} \times \text{Even} = \text{Even}$.
• Scenario 2 (a is Odd, b is Odd): $\text{a} \times \text{b} = \text{Odd} \times \text{Odd} = \text{Odd}$.

Since $\text{ab}$ can be odd (e.g., if $\text{a}=3, \text{b}=1$, then $\text{a – b}=2$ (even), but $\text{ab}=3$ (odd)), this expression is not always even.

Option 2: $\text{a}^{2} + \text{b}^{2} + 1$

• Scenario 1 (a is Even, b is Even): $\text{a}^{2} + \text{b}^{2} + 1 = (\text{Even})^{2} + (\text{Even})^{2} + 1 = \text{Even} + \text{Even} + 1 = \text{Even} + 1 = \text{Odd}$.

Since this expression can be odd, it is not always even. There is no need to check Scenario 2.

Option 3: $\text{a}^{2} + \text{b} + 1$

• Scenario 1 (a is Even, b is Even): $\text{a}^{2} + \text{b} + 1 = (\text{Even})^{2} + \text{Even} + 1 = \text{Even} + \text{Even} + 1 = \text{Even} + 1 = \text{Odd}$.

Since this expression can be odd, it is not always even. There is no need to check Scenario 2.

Option 4: $\text{ab – b}$

We can factor out 'b' from this expression: $\text{ab – b} = \text{b}(\text{a – 1})$.

• Scenario 1 (a is Even, b is Even):
  • Since 'a' is even, $\text{a – 1}$ will be odd.
  • 'b' is even.
  • So, $\text{b}(\text{a – 1}) = \text{Even} \times \text{Odd} = \text{Even}$.
• Scenario 2 (a is Odd, b is Odd):
  • Since 'a' is odd, $\text{a – 1}$ will be even.
  • 'b' is odd.
  • So, $\text{b}(\text{a – 1}) = \text{Odd} \times \text{Even} = \text{Even}$.

In both scenarios, the expression $\text{ab – b}$ results in an even number. Therefore, this expression must always be even.

Summary Table of Options

Based on our analysis, only the expression $\text{ab – b}$ consistently yields an even number under the given condition that $\text{a – b}$ is even.