We will prove or disprove the given statement:

Claim: For integers $a$ and $b$, if $a \mid b$, then $a^2 \mid b^2$.

Proof Attempt

We are given that $a \mid b$, which means there exists some integer $k$ such that:

$b = a k$

Squaring both sides:

$b^2 = (a k)^2 = a^2 k^2$

Since $b^2$ can be expressed as $a^2$ times an integer ($k^2$), this shows that $a^2 \mid b^2$.

Thus, the statement is true.

Conclusion

The statement is true for all integers $a$ and $b$. If $a$ divides $b$, then $a^2$ must divide $b^2$.

Would you like a more detailed explanation or examples? Here are some related questions to consider:

1. Can we generalize this result for higher powers, i.e., if $a \mid b$, does $a^n \mid b^n$ for any $n$?
2. What happens if $a$ is negative? Does the proof still hold?
3. Can we prove this using prime factorization?
4. What happens if we take the converse: If $a^2 \mid b^2$, does $a \mid b$ necessarily follow?
5. How does this property apply in modular arithmetic?

Tip: Squaring often preserves divisibility relationships, but always check for counterexamples when making general claims.