let n(> 1)be a positive integer, then the largest integer m such that (n^m + 1) divides $\left(1+\mathrm{n}+{\mathrm{n}}^2+\dots +{\mathrm{n}}^{127}\right)$ is 

${\mathrm{n}}^{\mathrm{m}}+1\text{ divides }1+\mathrm{n}+{\mathrm{n}}^2+\dots +{\mathrm{n}}^{127}$

therefore. $\frac{1+\mathrm{n}+{\mathrm{n}}^2+\dots +{\mathrm{n}}^{127}}{{\mathrm{n}}^{\mathrm{m}}+1}$ is an integer.

$\Rightarrow \frac{1-{\mathrm{n}}^{128}}{1-\mathrm{n}}\times \frac{1}{{\mathrm{n}}^{\mathrm{m}}+1}$ is an integer.

$\Rightarrow \frac{\left(1-{\mathrm{n}}^{64}\right)\left(1+{\mathrm{n}}^{64}\right)}{(1-\mathrm{n})\left({\mathrm{n}}^{\mathrm{m}}+1\right)}$ is an integer when largest

                            m = 64