I am confused with the below notations .

I know that

($a \equiv b \mod {n} )\iff ( n|(a-b)$ )

but what the below notation says ?

$a = b \mod {n}$

and in theorem 16 in this ,it's given as below

if $g \in\mathbb{Z_n^*} , r_1,r_2 \in \mathbb{Z_n}$ and $m_1,m_2 \in \mathbb{Z_n}$

($g^{m_1}r_1^n =g^{m_2}r_2^n \mod {n^2}) \implies (g^{m_1-m_2}r_1^n=r_2^n \mod{n^2}$)

In the above equation , both sides are divided with $g^{m_2}$

When we can divide both sides of a modular expression as in the above equation with a number ?

  • $\begingroup$ This is wrong, $a\equiv b\mod n\iff n|(a-b)$ not the other way around. $\endgroup$ – Adam Hughes Jul 2 '14 at 17:29
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    $\begingroup$ When the number we are dividing by is relatively prime to the modulus. $\endgroup$ – André Nicolas Jul 2 '14 at 17:31
  • $\begingroup$ @AdamHughes yeah , edited . $\endgroup$ – hanugm Jul 2 '14 at 17:31
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    $\begingroup$ You can use the Euclidean algorithm to see that "division" is legal when $gcd(g,n)=1$. $\endgroup$ – Adam Hughes Jul 2 '14 at 17:33
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    $\begingroup$ @hanu $\ a = b\ {\rm mod}\ n\ $ could mean either $\ a\equiv b\pmod n\ $ or $\ a = (b\ {\rm mod}\ n),\,$ i.e. the former combined with $\,0\le a < n.\ \ $ $\endgroup$ – Bill Dubuque Jul 2 '14 at 17:45

$$g^{m_1}r_1^{n} \equiv g^{m_2}r_2^{n} \pmod {n^2} \Rightarrow n^2 \mid g^{m_1}r_1^{n} - g^{m_2}r_2^{n} \\ \Rightarrow n^2 \mid g^{-m_2} \cdot (g^{m_1}r_1^{n} - g^{m_2}r_2^{n}) \Rightarrow n^2 \mid g^{m_1-m_2}r_1^n-r_2^n$$

EDIT: We can multiply with $g^{-m_2}$,because, we know that $g \in \mathbb{Z}^*$,so it is a unit,therefore $g^{-1}$ exists.

In general, if $m \mid a-b \Rightarrow m \mid x(a-b), \forall x \in \mathbb{Z}$

  • $\begingroup$ but is $g^{-m_2} \in \mathbb{Z}$ ? $\endgroup$ – hanugm Jul 2 '14 at 17:39
  • $\begingroup$ I edited my answer..Tell me if you understand it now!!! $\endgroup$ – evinda Jul 2 '14 at 17:41
  • $\begingroup$ @hanu: an inverse can be found using the Euclidean algorithm, as I indicated in my previous comment. $\endgroup$ – Adam Hughes Jul 2 '14 at 17:43
  • $\begingroup$ Yeah , understood . what about the notation $a=b \mod {n}$ , is it same as $a \equiv b \mod{n}$ $\endgroup$ – hanugm Jul 2 '14 at 17:45
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    $\begingroup$ Nice! it is the same,but it is better to write it like that : $$a \equiv b \pmod n$$ because it isn't exactly an equality!!!We are working $\mod n$.. $\endgroup$ – evinda Jul 2 '14 at 17:47

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