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$ax\equiv 1 \bmod n$ is stated to have a solution if and only if $(a,n)=1$.
Based on this have a few questions:
(i) This means, $\exists b \in \mathbb {Z}, ax-1= bn \implies (ax-bn)=1$.
(ii) Can the last equality be viewed as a variant of the usual equality : $ax + by =1$. If so, what is the logic.
In terms of the geometrical interpretation , it is having an opposite slope for the straight line.

(iii) How can I interpret the equality $a\equiv 1 \bmod n$, if valid. I mean the difference from the title equality, if valid, by removal of variable $x$.

jiten
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  • @mathlove So, it means that the equality $\exists a,b, n, \in \mathbb{Z}, ax\equiv b \bmod n$ is a super-set of the usually seen equality: $a\equiv b \bmod n$. If so, why this very important fact is not stated in any book on elementary number theory. – jiten Mar 04 '18 at 05:39
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    For (i)(ii), note that 1) "$(a,n)=1$", 2) "There exist integers $x,b$ such that $ax+bn=1$" and 3) "$ax\equiv 1\pmod n$ has a solution" are equivalent. – mathlove Mar 04 '18 at 05:48

1 Answers1

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(i)$$ax \equiv 1 \pmod{n}$$ does mean $\exists b \in \mathbb{Z}$, such that $$ax-1 = bn.$$

(ii) If you are comparing with a linear diophantine equation $$ax+\color{blue}{B}\color{green}y = 1$$ where we know that is has a solution iff $\gcd(a,B)=1$, then the right comparison should be $$ax+\color{blue}n\color{green}{(-b)}=1$$

Note that $a$ and $n$ are given.

If you dislike $-b$ explicitly, you can let $c=-b$ and we have $$ax+\color{blue}n\color{green}{c}=1$$

(iii) hmm... it means $a-1$ is a multiple of $n$. If you started from $$ax\equiv 1 \pmod{n}$$ and you found that $a\equiv 1 \pmod{n}$, then we know that $x \equiv 1 \pmod{n}$ is a solution to the equation.

Remark:

$\exists b \in \color{red}{\exists} Z$, I don't quite understand the second exists statement.

Siong Thye Goh
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  • Thanks for the answer. Have corrected the error for $\exists Z$ – jiten Mar 04 '18 at 06:49
  • I request elaboration on the third part's answer by you. I am confused by the implication: Given $ax\equiv 1 \pmod n$, then $a\equiv 1 \pmod n \implies x\equiv 1 \pmod n$. It is confusing as need to know if the above implication is 'always' true. – jiten Mar 04 '18 at 07:43
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    I mean $ax \equiv 1 \pmod{n}$ and $a \equiv a(1) \equiv 1 \pmod{n}$ implies that $x \equiv 1 \pmod{n}$ is a solution to $ax \equiv 1 \pmod{n}$. – Siong Thye Goh Mar 04 '18 at 07:50
  • Please tell the meaning of $a(1)$, as is new to me. – jiten Mar 04 '18 at 08:03
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    it means $a$ times $1$. – Siong Thye Goh Mar 04 '18 at 08:10
  • But, why it was used in equivalence relations by your comment, if it is the same as $a$. – jiten Mar 04 '18 at 08:14
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    I suspect i have misunderstood your quesiton $(iii)$, initially I interpreted the question as what can we conclude if we are given $a \equiv 1 \pmod{n}$ and $ax \equiv 1 \pmod{n}$. The difference between the two is for the first one it is $ax -1$ is multiple of $n$ and the second one means $a-1$ is a multiple of $n$. – Siong Thye Goh Mar 04 '18 at 08:21
  • Kindly look at my post at : https://math.stackexchange.com/q/2678878/424260, and kindly help with understanding how the final step of induction is used, and how it is strong induction. – jiten Mar 06 '18 at 06:28