Counting

We can solve this problem by using a counting method. Let's consider the number of ways we can color the five squares if we don't have any restrictions on the colors. There are three choices for the color of the first square, and three choices for the color of the second square, and so on. Therefore, there are a total of 3 x 3 x 3 x 3 x 3 = 3^5 = 243 ways to color the five squares without any restrictions.

Now, we need to subtract the number of ways that violate the condition that no two consecutive squares have the same color. Let's call this condition the "no-consecutive" condition.

If the first square is red, then we can color the second square in two ways (yellow or blue). For each choice of the second square, there is only one choice for the third square (the color that is different from the second square). Similarly, there is only one choice for the fourth square, and two choices for the fifth square. Therefore, there are 2 x 1 x 1 x 2 = 4 ways to color the five squares if the first square is red and we satisfy the no-consecutive condition.

If the first square is not red, then we can color it in two ways (yellow or blue). For each choice of the first square, there are two choices for the second square (the color that is different from the first square). For each choice of the second square, there is only one choice for the third square (the color that is different from the second square). Similarly, there is only one choice for the fourth square, and two choices for the fifth square. Therefore, there are 2 x 2 x 1 x 1 x 2 = 8 ways to color the five squares if the first square is not red and we satisfy the no-consecutive condition.

Therefore, the total number of ways to color the five squares that satisfy the no-consecutive condition is 4 + 8 = 12.

However, we also need to ensure that at least three squares are red. There are three cases to consider:

- Exactly three squares are red: There are 3 ways to choose which three squares will be red, and 2 choices for each non-red square. Therefore, there are 3 x 2 x 2 x 2 = 24 ways to color the squares in this case.
- Exactly four squares are red: There are 5 ways to choose which four squares will be red, and 2 choices for the non-red square. Therefore, there are 5 x 2 = 10 ways to color the squares in this case.
- All five squares are red: There is only 1 way to color the squares in this case.

Therefore, the total number of ways to color the squares with at least three red squares and satisfying the no-consecutive condition is 24 + 10 + 1 = 35.

So the final answer is 35 ways.

We can solve this problem using casework, based on the number of red squares.

Case 1: Three red squares If we have three red squares, then the other two squares must be different colors. We can choose the colors for these two squares in 2 * 2=4 ways (since we have two choices for the first square, and two choices for the second square). We can then arrange the five squares in 5!/(3!2!)=10 ways (since there are 3 identical red squares and 2 identical squares of the other color). Therefore, there are 4 * 10=40 ways to color the squares in this case.

 

Case 2: Four red squares If we have four red squares, then the remaining square must be a different color. We can choose the color for this square in 2 ways, and we can arrange the five squares in 5!/(4!1!)=5 ways. Therefore, there are 2*5=10 ways to color the squares in this case.

 

Case 3: Five red squares If we have five red squares, then there are no choices for the other colors. There is only one way to arrange the five squares.

 

Therefore, the total number of ways to color the squares is 40+10+1=51.

Answer: There are 51 ways to color the five squares.