Question

If m and n are any two odd positive integers with n < m, then the largest positive integer which divides all numbers of the form, $m^2-n^2$ can be

• A. 4
• B. 6
• C. 8
• D. 9

Answer 1

The correct option is C 8

$Letm=2k-1andn=2P-1,p<k$
$Then{m}^2-n^2=(m+n)(m-n)$
Further if k and p both even, then k-p is even but k+p-1 is odd
If k and p both odd then k-p is even but k+p-1 is odd. If one is even and other odd then k-p is odd but k+p-1 is even. Thus in every case (k-p)(k+p-1) even
$\therefore m^2-n^2$ is divisible by $4\times 2=8$. Hence, $m^2-n^2$ is divisible by 8 or any multiple of 8. The largest integer among the given options is 8