Solving an alphametic puzzle by eavesdropping

Question

I took part in a meeting of puzzle maniacs. I overheard a conversation between two persons.

A: Hey, solve this alphametic problem. LHS is a square of two-digit number and RHS is a four-digit number.

B: Does it contain some middle steps?

A: No. That's a whole. Can you solve it?

After some minutes, B asked.

B: The answer cannot be determined. If we know whether the number is odd or even, it could be unique.

A: Wait, please. Let me read again. Oh, sorry. The parity was mentioned in the problem.

A read the sentence about parity to B.

B: OK. That figures. If the parity was given reversely, we couldn't get the unique answer.

I did not see their problem. But, I could solve it just by eavesdropping.

What's the answer of this alphametic problem?

PS. The parity in the dialogue means whether the RHS number is odd or even.

Answer 1

The answer is:

$93^2 = 8649$

Reasoning:

There are 3 possible puzzles matching the conversation:
$(AB)^2 = CDCB$, where AB can be 56, 45, 81 or 91
$(AB)^2 = CDDB$, where AB can be 46, 35, 65 or 85
$(AB)^2 = CDEA$, where AB can be 42, 48 or 93
If A said that the number is even, and B deduced the unique answer, then AB can be either 56 or 46, leaving "I" unable to solve it.
Therefore the number is odd, and the answer is $93^2 = 8649$.

Earlier answer (before the edit):

How about:

Puzzle: $(AB)^2 = CDEA$
The solutions are: $42^2 = 1764$, $48^2 = 2304$, $93^2 = 8649$

For the parity part:

If we know A is odd, AB is 93, otherwise its 42 or 48
If we know B is odd, AB is 93, otherwise its 42 or 48
If we know C is odd, AB is 42, otherwise its 48 or 93
If we know D is even, AB is 93, otherwise its 42 or 48
E must be even in this puzzle

However...

We still can't deduce the answer of the alphametic problem, unless we heard the sentence about parity clearly. If we know which number/letter's parity is given, we can get the answer.

How I get the answer:

I listed all the squares between 1000~9999. There are only 67 of them so its relatively easy to bruteforce. After listing the squares I looked for the potential puzzles with 3 solutions or more and find this.

Answer 2

I have no idea how to prove its uniqueness, but here is a possible situation:

Puzzle: $AB^2=CDCB$.
The solutions: $56^2=3136$, $81^2=6561$, $91^2=8281$, $45^2=2025$.
Unique if we know the numbers are even.

How I get this:

Let the last digit be same (so my work will be easier) and list squares of two-digit integers ending with $1,5,6$.