Let $n( 1)$ be a positive integer then the largest integer $m$ such that $(n^{m} + 1)$ divides $(1 + n + n^{2} + ... + n^{127})$, is

Question

Let $n( > 1)$ be a positive integer then the largest integer $m$ such that $(n^{m} + 1)$ divides $(1 + n + n^{2} + ... + n^{127})$, is

Dr. Abubacker · Accepted Answer

(c)
Since, $n^{m} + 1$ divides $1 + n + n^{2} + ... + n^{127}$
Therefore, $\frac{1 + n + n^{2} + ... + n^{127}}{n^{m} + 1}$is an integer.
$\Rightarrow \ \frac{1 - n^{128}}{1 - n} 	imes \frac{1}{n^{m} + 1}$ is an integer.
$\Rightarrow \ \frac{\left( 1 - n^{64} ight)(1 + n^{64})}{(1 - n)(n^{m} + 1)}$ is an integer when largest
$m = 64$

Q. Let $n( > 1)$ be a positive integer then the largest integer $m$ such that $(n^{m} + 1)$ divides $(1 + n + n^{2} + ... + n^{127})$, is

Explanation

Reviewed by Dr. Abubacker

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