Let b be a positive integer and `a=b^(2)-b`. If `bge4` then `a^(2)-2a` is divisible by?

Options

A. 15
B. 20
C. 24
D. All of the above

Answer

To solve the problem, we start with the expression for \( a \):

1. **Define \( a \)**:
\[ a = b^2 - b \]

2. **Calculate \( a^2 - 2a \)**:
We need to compute \( a^2 - 2a \).
\[ a^2 = (b^2 - b)^2 = b^4 - 2b^3 + b^2 \]
\[ 2a = 2(b^2 - b) = 2b^2 - 2b \]
Now, substitute these into the expression:
\[ a^2 - 2a = (b^4 - 2b^3 + b^2) - (2b^2 - 2b) \]
Simplifying this:
\[ a^2 - 2a = b^4 - 2b^3 + b^2 - 2b^2 + 2b \]
\[ = b^4 - 2b^3 - b^2 + 2b \]

3. **Factor \( a^2 - 2a \)**:
We can rearrange and factor the expression:
\[ a^2 - 2a = b^4 - 2b^3 - b^2 + 2b \]
To factor this, we can group terms:
\[ = b^4 - 2b^3 + 2b - b^2 \]
\[ = b^4 - 2b^3 + b^2 + 2b - 3b^2 \]
\[ = (b^4 - 2b^3 + b^2) + (2b - 3b^2) \]

4. **Check divisibility**:
Now, we need to check if \( a^2 - 2a \) is divisible by any specific integers for \( b \geq 4 \).

Let's evaluate \( a^2 - 2a \) for specific values of \( b \):
- For \( b = 4 \):
\[ a = 4^2 - 4 = 16 - 4 = 12 \]
\[ a^2 - 2a = 12^2 - 2 \cdot 12 = 144 - 24 = 120 \]
\( 120 \) is divisible by \( 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 \).

- For \( b = 5 \):
\[ a = 5^2 - 5 = 25 - 5 = 20 \]
\[ a^2 - 2a = 20^2 - 2 \cdot 20 = 400 - 40 = 360 \]
\( 360 \) is divisible by \( 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 60, 72, 90, 120, 180, 360 \).

- For \( b = 6 \):
\[ a = 6^2 - 6 = 36 - 6 = 30 \]
\[ a^2 - 2a = 30^2 - 2 \cdot 30 = 900 - 60 = 840 \]
\( 840 \) is divisible by \( 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 42, 60, 70, 84, 105, 120, 140, 210, 280, 420, 840 \).

From these calculations, we can observe that \( a^2 - 2a \) is divisible by \( 120 \) for \( b = 4 \) and is divisible by \( 360 \) for \( b = 5 \) and \( 840 \) for \( b = 6 \).

Thus, we can conclude that:

**Final Answer**: \( a^2 - 2a \) is divisible by \( 120 \) for \( b \geq 4 \).