To prove by contradiction that there is no integer $n$ such that $n^2$ is even and $n^3$ is odd, we proceed step by step.

Proof by Contradiction:

1. Assume the opposite: Suppose there exists an integer $n$ such that $n^2$ is even and $n^3$ is odd.

2. Analyze the parity of $n$:

   • If $n^2$ is even, then $n$ must be even.
     • Why? If $n$ were odd, then $n^2 = n \times n$ would also be odd (since the product of two odd numbers is odd). Hence, for $n^2$ to be even, $n$ must be even.

3. Implication for $n^3$:

   • If $n$ is even, then $n^3 = n \times n \times n$ must also be even.
     • Why? The cube of an even number is always even (since multiplying any even number by itself multiple times results in an even number).

4. Contradiction: From Step 3, $n^3$ must be even. But this contradicts the assumption that $n^3$ is odd.

Conclusion:

Since assuming the existence of such an integer $n$ leads to a contradiction, it must be impossible for $n^2$ to be even and $n^3$ to be odd. Therefore, no such integer $n$ exists.

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Would you like more details or have any questions? Here are 5 related questions for deeper understanding:

1. Can you prove by contradiction that the square of an odd number is always odd?
2. Why does the cube of an even number always remain even?
3. What other properties of numbers can we prove using contradiction?
4. How would the proof differ if we tried proving the opposite statement, $n^2$ odd and $n^3$ even?
5. Can this proof be adapted for rational numbers instead of integers?

Tip: When proving by contradiction, assume the opposite of the desired conclusion and work logically until you reach a contradiction.