Divisibility
Let $n\geq 2$ and $k$ be positive integers. Prove that $(n-1)^{2} \mid (n^{k}-1)$ if and only if $(n-1) \mid k$.
"Questions we can't answer are far better than answers we can't question"
Re: Divisibility
$n^{k}-1=(n-1)(n^{k-1}+n^{k-2}+......+n^{0})$
so, we need to prove that $(n-1)$ divides $(n^{k-1}+n^{k-2}+......+n^{0})$
now,
$n^{k-1}\equiv 1(modn-1)$
$n^{k-2}\equiv 1(modn-1)$
.
.
.
$n^{0}\equiv 1(modn-1)$
so , $(n^{k-1}+n^{k-2}+......+n^{0})\equiv k\equiv 0(modn-1)$
so, we need to prove that $(n-1)$ divides $(n^{k-1}+n^{k-2}+......+n^{0})$
now,
$n^{k-1}\equiv 1(modn-1)$
$n^{k-2}\equiv 1(modn-1)$
.
.
.
$n^{0}\equiv 1(modn-1)$
so , $(n^{k-1}+n^{k-2}+......+n^{0})\equiv k\equiv 0(modn-1)$
Re: Divisibility
Same as my proof.Tahmid wrote:$n^{k}-1=(n-1)(n^{k-1}+n^{k-2}+......+n^{0})$
so, we need to prove that $(n-1)$ divides $(n^{k-1}+n^{k-2}+......+n^{0})$
now,
$n^{k-1}\equiv 1(modn-1)$
$n^{k-2}\equiv 1(modn-1)$
.
.
.
$n^{0}\equiv 1(modn-1)$
so , $(n^{k-1}+n^{k-2}+......+n^{0})\equiv k\equiv 0(modn-1)$
"Questions we can't answer are far better than answers we can't question"