Let the three consecutive integers be represented as $x - 1$, $x$, and $x + 1$.

The sum of the squares of these integers is given as 110, so we set up the equation:

$(x - 1)^2 + x^2 + (x + 1)^2 = 110$

Expanding each square term, we get:

$(x - 1)^2 = x^2 - 2x + 1$ $x^2 = x^2$ $(x + 1)^2 = x^2 + 2x + 1$

Substituting these into the equation, we have:

$(x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 110$

Simplify by combining like terms:

$3x^2 + 2 = 110$

Subtract 2 from both sides:

$3x^2 = 108$

Now, divide by 3:

$x^2 = 36$

Take the square root of both sides:

$x = 6 \quad \text{or} \quad x = -6$

Case 1: $x = 6$

The three consecutive integers are $5$, $6$, and $7$.

Case 2: $x = -6$

The three consecutive integers are $-7$, $-6$, and $-5$.

Thus, the two possible sets of integers are:

1. $5, 6, 7$
2. $-7, -6, -5$

Would you like further details on how this was derived or have any questions?

Here are some related questions:

1. How do you generalize finding sums of squares for any number of consecutive integers?
2. What if we had consecutive even or odd integers instead?
3. How does the sum of cubes of consecutive integers differ from the sum of squares?
4. What other types of equations can result in finding consecutive integers?
5. How does this method change if fractions are allowed?

Tip: Consecutive integer problems often lead to quadratic equations. Solving these requires familiarity with factoring and square root properties.