Mathematics - Binomial Theorem Question with Solution | TestHub

Let $m=2k-1$ and $n=2p-1,p<k$ Then $m^2-n^2=(m+n)(m-n)$ $=(2k+2p-2)(2k-2p)=4(k+p-1)(k-p)$ Further if k and p both even, then $\mathrm{k}-\mathrm{p}$ is even but $\mathrm{k}+\mathrm{p}-1$ 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 $\mathrm{k}+\mathrm{p}-1$ 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