Question

Show that if m and n are both square numbers, then mn is a square number

Answer 1

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.