Let \( m \) and \( n \) be square numbers. By definition, a square number can be expressed as the square of an integer.
We can write: \[ m = a^2 \quad \text{for some integer } a \] \[ n = b^2 \quad \text{for some integer } b \]
Now, we multiply \( m \) and \( n \): \[ mn = m \cdot n = a^2 \cdot b^2 \]
Using the properties of exponents, we can combine the terms: \[ mn = (ab)^2 \]
Here, \( ab \) is also an integer since both \( a \) and \( b \) are integers. Therefore, \( (ab)^2 \) is a square number.
Thus, we have shown that the product \( mn \) is a square number. Therefore, if \( m \) and \( n \) are both square numbers, then \( mn \) is also a square number. This completes the proof.