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If $m$ is a natural number greater than $1$ and is not prime, then we know that $m=ab$, where $1<a<m$ and $1<b<m$. Show that there is no integer $x$ such that $ax≡1\pmod{m}$.

My thoughts:

What I think is $ax$ can only congruent to multiple of $a$ under mod $a\cdot b$ and since $a>1$ that means it could never congruent to $1$.

\begin{align} ax&\equiv y\pmod{a\cdot b}\tag*{Let $x\in\mathbb{Z}$, solve $y$}\\ y&=a(b\cdot k+x)\tag*{Case1: $ax<ab;k\in\mathbb{Z}$}\\ y&=a(x-b+b\cdot k)\tag*{Case2: $ax\ge ab;k\in\mathbb{Z}$}\\ y&\neq 1\tag*{Since $a>1$}\\ \end{align}

Is this idea correct, are there other approaches to prove this statement ?

Ethan
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    There's no need for two cases. Instead we have $$ \exists, x!:\ a,x\equiv c!!!\pmod{!ab},\Rightarrow, \exists, x,y!:\ \color{#c00}a,x+\color{#c00}ab,y = c,\Rightarrow, \color{#c00}a\mid c,\ \ {\rm so}\ \ |c|\ge |a| > 1,\Rightarrow, c\neq 1$$ – Bill Dubuque Jan 20 '20 at 04:59
  • @BillDubuque thx, indeed a better approache, I should notice $a\mid c$ implies $|c|\ge |a|$ – Ethan Jan 20 '20 at 05:25
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    I'd rewrite your proof as follows: if $,y\equiv ax\pmod{!ab},$ then, by definition of congruence, there is an integer $k$ with $, y = ax+kab = a(x+kb)\ $ so $\ a\mid y,,$ so $\ |y|>|a|\ge 1\ \ $ – Bill Dubuque Jan 20 '20 at 05:31
  • i.e. any common factor $,a,$ a residue $r$ has with a modulus $n$ persists as a factor in every $\ r' \equiv r\pmod{!n}\ $ Such a nontrivial factor exists $\iff r,$ is not invertible $\bmod n,,$ which is independent of the residue rep chosen. – Bill Dubuque Jan 20 '20 at 05:35

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