Let the two consecutive odd integers be $x$ and $x + 2$. According to the problem, the sum of their squares is 34: $x^2 + (x + 2)^2 = 34$

First, expand the equation: $x^2 + (x + 2)^2 = x^2 + x^2 + 4x + 4 = 34$

Combine like terms: $2x^2 + 4x + 4 = 34$

Subtract 34 from both sides to set the equation to zero: $2x^2 + 4x + 4 - 34 = 0$ $2x^2 + 4x - 30 = 0$

Divide the entire equation by 2 to simplify: $x^2 + 2x - 15 = 0$

Next, factor the quadratic equation: $x^2 + 2x - 15 = (x + 5)(x - 3) = 0$

Set each factor equal to zero: $x + 5 = 0 \quad \text{or} \quad x - 3 = 0$ $x = -5 \quad \text{or} \quad x = 3$

Thus, the two pairs of consecutive odd integers are:

1. $x = -5$ and $x + 2 = -3$
2. $x = 3$ and $x + 2 = 5$

So, the pairs of consecutive odd integers whose squares sum to 34 are $\boxed{-5 \text{ and } -3}$ or $\boxed{3 \text{ and } 5}$.

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Here are 8 related questions that expand on this information:

1. How can you determine if an equation can be factored easily?
2. What are the steps to solving a quadratic equation by factoring?
3. What are other methods to solve quadratic equations besides factoring?
4. How do you recognize and solve problems involving consecutive integers?
5. Can you explain the relationship between the roots and coefficients of a quadratic equation?
6. How do you verify if a solution to a quadratic equation is correct?
7. What are the characteristics of odd and even integers?
8. How can you apply the quadratic formula to solve quadratic equations?

Tip: When working with consecutive integers, setting up the problem correctly with the right expressions for the integers is crucial for simplifying the process and avoiding mistakes.