Question

Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$. The largest integer which divides all possible numbers of the form $m^2-n^2$ is

• A. $2$
• B. $4$
• C. $6$
• D. $8$
• E. $16$

Answer 1

The correct option is D $8$

$m=2r+1$ where $r=0,\pm 1,\pm 2,....;$
$n=2s+1$ where $s=0,\pm 1,\pm 2,.....;$
$m^2-n^2=4r^2+4r+1-4s^2-4s-1$
$=4(r-s)(r+s+1)$, a number certainly divisible by $4$.
If $r$ and $s$ are both even or both odd, $r-s$ is divisible by $2$, and $r+s+1$ is not.
If $r$ and $s$ one even and one odd, then $r+s+1$ is divisible by $2$, and $r-s$ is not. Thus $m^2-n^2$ is divisible by $4\cdot 2=8$.