If  m and n are positive integers and (m -n) is an even number, then (m^2 – n^2) will always be divisible by:

Given:
m and n are positive integers and (m - n) is an even number.

This means m and n have the same parity, i.e., both are even or both are odd.

We know the identity:

$(m^2 - n^2) = (m - n)(m + n)$

Case 1: Both $m$ and $n$ are even

Let $m=2a$ and $n=2b$

Then
$m^2 - n^2 = (2a)^2 - (2b)^2$
$= 4a^2 - 4b^2$
$= 4(a^2 - b^2)$

So it is divisible by 4.

Case 2: Both $m$ and $n$ are odd

Let $m=2a+1$ and $n=2b+1$

Then
$m^2 = (2a+1)^2 = 4a^2 + 4a + 1$
$n^2 = (2b+1)^2 = 4b^2 + 4b + 1$

$m^2 - n^2 = (4a^2 + 4a + 1) - (4b^2 + 4b + 1)$
$= 4(a^2 + a - b^2 - b)$

So it is again divisible by 4.

Therefore, in both cases $(m^2 - n^2)$ is divisible by 4.

✅ Correct Answer: 4