. Suppose a,b ∈ Z. If a^2 (b^2 −2b) is odd, then a and b are odd. ﻿ can you check my proof: so i will assume the contrapositive and thus we have: if a OR b is EVEN, then a^2(b^2 - 2b) is even. ﻿ assume b is even so b = 2k: a^2 (b^2 - 2b) = a^2 (4k^2 - 4k) ﻿ a can be either even or odd so: if a is even: a = 2m ﻿ a^2 (b^2 - 2b) = 4m^2 ( 4k^2 - 4k) which is even ﻿ if a is odd: a = 2n + 1 and you can do the same thing here which will prove to be even ﻿ therefore we proed the contrapositive and that proves the original preposition ﻿

Your proof is mostly correct in structure, but it could use a bit more clarity and precision, especially in the case when \( a \) is odd. Let's go through the proof step by step and refine it.

### Problem restatement:
We are asked to prove that if \( a^2(b^2 - 2b) \) is odd, then both \( a \) and \( b \) must be odd. 

### Proof by Contrapositive:
The contrapositive of the statement is:  
*If \( a \) or \( b \) is even, then \( a^2(b^2 - 2b) \) is even.*

If we can prove the contrapositive, the original statement will follow.

#### Case 1: Assume \( b \) is even
Let \( b = 2k \) for some integer \( k \). Now, let's compute \( a^2(b^2 - 2b) \):

\[
b^2 - 2b = (2k)^2 - 2(2k) = 4k^2 - 4k
\]
Thus:
\[
a^2(b^2 - 2b) = a^2(4k^2 - 4k) = 4a^2(k^2 - k)
\]
This expression is clearly divisible by 4, which means \( a^2(b^2 - 2b) \) is even regardless of whether \( a \) is odd or even.

#### Case 2: Assume \( a \) is even
Let \( a = 2m \) for some integer \( m \). In this case, we don't need to assume anything about \( b \); we just compute \( a^2(b^2 - 2b) \):

\[
a^2 = (2m)^2 = 4m^2
\]
Thus:
\[
a^2(b^2 - 2b) = 4m^2(b^2 - 2b)
\]
This expression is clearly divisible by 4, which means \( a^2(b^2 - 2b) \) is even regardless of whether \( b \) is odd or even.

### Conclusion:
In both cases (when either \( a \) or \( b \) is even), we have shown that \( a^2(b^2 - 2b) \) is even. Therefore, the contrapositive is true, which implies that the original statement is true as well: *If \( a^2(b^2 - 2b) \) is odd, then both \( a \) and \( b \) must be odd.*

### Improvements:
- When handling the case where \( a \) is odd, you should explicitly show that when \( a = 2n + 1 \), the expression still leads to an even result when \( b \) is even, though your intention was clear.

Would you like a more detailed breakdown of the odd case or further clarifications? 

Here are five follow-up questions to deepen your understanding:

1. How would the proof change if the statement only involved \( a \), not \( b \)?
2. Can you generalize this proof to more complex expressions involving odd and even variables?
3. How can you verify if an expression involving powers and products of integers is odd or even without direct calculation?
4. What does this proof tell you about the interaction between parity (odd/even nature) in mathematical expressions?
5. How would the proof differ if you were asked to prove this directly rather than via the contrapositive?

**Tip:** When dealing with odd and even proofs, it is useful to explicitly substitute \( a = 2m \) or \( a = 2m + 1 \) at each step to avoid confusion and ensure clarity.

Suppose a, b ∈ Z. If a^2 (b^2 − 2b) is odd, then a and b are odd. Can you check my proof: so I will assume the contrapositive and thus we have: if a OR b is EVEN, then a^2(b^2 - 2b) is even. Assume b is even so b = 2k: a^2 (b^2 - 2b) = a^2 (4k^2 - 4k). A can be either even or odd so: if a is even: a = 2m. a^2 (b^2 - 2b) = 4m^2 (4k^2 - 4k) which is even. If a is odd: a = 2n + 1 and you can do the same thing here which will prove to be even. Therefore we proved the contrapositive and that proves the original proposition.

This detailed proof verifies that if a^2(b^2 - 2b) is odd, then both a and b must be odd. It demonstrates the use of contrapositive proof methods and explores properties of even and odd integers, making it suitable for advanced high school or undergraduate math students.

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