Question

Let n>1, be a positive integer. then the largest number m, such that ( nm + 1) divides ( 1 + n + n2 + n3 + n4 +..........n127 ) is:

Answer 1

Question:

Let $n>1$ be a positive integer. Then the largest number $m$ such that $n^m+1$ divides $1+n+n^2+n^3+n^4+\ \dots\ +n^{127}$ is... ?

Solution:

The series,

$1+n+n^2+n^3+n^4+\ \dots\ +n^{127},$

consisting of 128 terms, is a geometric series with $a=1\ \&\ r=n>1$

So the sum is given by,

$\dfrac {a(r^{128}-1)}{r-1}=\dfrac {n^{128}-1}{n-1}$

But,

$n^{128}-1=(n^{64}+1)(n^{32}+1)(n^{16}+1)(n^8+1)(n^4+1)(n^2+1)(n+1)(n-1)$

Hence the sum is,

$\dfrac {n^{128}-1}{n-1}=(n^{64}+1)(n^{32}+1)(n^{16}+1)(n^8+1)(n^4+1)(n^2+1)(n+1)$

Given that this sum is divisible by $n^m+1$ where $m$ has the greatest value.

While comparing the sum factorised with $n^m+1,$ we can see that,

$\large\boxed {\boxed {\max(m)=\mathbf{64}}}$

Hence 64 is the answer.