The problem states:

Given that $a$ and $b$ are positive odd integers, and $a^b b^a$ is a perfect square, prove that $ab$ is a perfect square.

Solution

1. Step 1: Since $a^b b^a$ is a perfect square, it means $a^b b^a = k^2$ for some integer $k$.

2. Step 2: Let’s analyze the powers of $a$ and $b$ in the expression $a^b b^a$. $a^b b^a = a^b \cdot b^a$ To be a perfect square, all exponents in the factorization of $a$ and $b$ must be even.

3. Step 3: Assume the prime factorizations: $a = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k} \quad \text{and} \quad b = q_1^{f_1} q_2^{f_2} \dots q_m^{f_m}$ Then: $a^b = p_1^{e_1 b} p_2^{e_2 b} \dots p_k^{e_k b}$ $b^a = q_1^{f_1 a} q_2^{f_2 a} \dots q_m^{f_m a}$

4. Step 4: Since $a^b b^a$ is a perfect square, each exponent in the prime factorization of $a$ and $b$ must be even. Therefore, $e_i b$ and $f_j a$ must be even for all $i$ and $j$.

5. Step 5: Because $a$ and $b$ are both odd, $e_i$ and $f_j$ must be even to satisfy the evenness of $e_i b$ and $f_j a$. Therefore, $a$ and $b$ are themselves perfect squares.

6. Step 6: Since $a$ and $b$ are perfect squares, their product $ab$ is also a perfect square.

Thus, we have proven that $ab$ is a perfect square.

Would you like further details or have questions?

Additional Questions

1. How does the parity (odd/even) of integers influence the solution?
2. Can this proof be generalized for other values of $a$ and $b$ (not just odd integers)?
3. Why does the factorization require even exponents to form a perfect square?
4. How do prime factorizations assist in proving statements about powers and products?
5. What would change if $a$ and $b$ were even integers?

Tip: When dealing with perfect squares in proofs, factorize each term to ensure that all exponents are even, which confirms the square nature.