Question

If $n>1$, then the values of the positive intetger $m$ for which $n^m+1$ divides $a=1+n+n^2+...+n^{63}$ is

• A. $8$
• B. $16$
• C. $32$
• D. $64$

Answer 1

Answer: (A), (B), (C)

$32$
$8$
$16$

We know that,

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots +b^{n-2}a+b^{n-1})\Rightarrow (a^{n-1}+a^{n-2}b+\dots +b^{n-2}a+b^{n-1})=\frac{a^n-b^n}{a-b}$

Given here,

$a=1+n+n^2+n^3+...+n^{63}=\frac{n^{64}-1}{n-1}=\frac{(n^{32}-1)(n^{32}+1)}{n-1}=\frac{(n^{16}-1)(n^{16}+1)(n^{32}+1)}{n-1}=\dots .=\frac{(n-1)(n+1)(n^2+1)(n^4+1)(n^8+1)(n^{16}+1)(n^{32}+1)}{n-1}=(n+1)(n^2+1)(n^4+1)(n^8+1)(n^{16}+1)(n^{32}+1)$

Hence, the values of the positive integer $m$ for which $n^m+1$ divides $a$ are : $1,2,4,8,16,32$