Prove a modular equation

Question

If $a\equiv 1 \pmod n$ and $d\mid n$, then $a\equiv1 \pmod d$ is true.

That is the question I am trying to understand so I have some understanding of the question but cant place my finger on exactly why it is true.

$n = d \times k$ for $k\in\mathbb{Z}$

$a = k \times n +1$ and $a= k \times d +1$

$n \mid(a-1)$ and $d \mid(a-1)$

If $(a-1)=m\times n$ and $n=d\times k$ then $(a-1)=d\times (m\times k)$. — lulu, Mar 18 '21 at 23:34
You said that $(a-1)\equiv 0 \pmod n$. that means $n,|,(a-1)$. Thus there is some integer $m$ such that $(a-1)=m\times n$. — lulu, Mar 18 '21 at 23:58
$a\equiv1\pmod n\iff a=1+nN$ so if $n=dm$ you have $a=1+dmN\Rightarrow a\equiv1\pmod d$ — Piquito, Mar 19 '21 at 00:24
Yes, congruences persist mod divisors of the modulus, as proved in the dupe. — Bill Dubuque, Mar 19 '21 at 01:28

score 0 · Accepted Answer · answered Mar 19 '21 at 00:00

0

If $a$ is congruent to 1 modulo $n$, then $a$ is one more than a multiple of $n$.

If $d$ divides $n$, then $n$ is a multiple of $d$. Hence, $a$ is also one more than a multiple of $d$. So we say $a$ is congruent to 1 module $d$.

answered Mar 19 '21 at 00:00

Dan Barry

can this be shown with an equation like what @lulu pointed out? – zellez11 Mar 19 '21 at 00:03
Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Mar 19 '21 at 01:28

1 Answers1