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Prove that if $n,m\in\mathbb Z^{+}$, then $\forall x,y \in \{0,1,2,…,m-1\} \land x\not=y$, $n(x-y)\not\equiv0\pmod m$ if and only if $\gcd(n,m)=1$.

This problem is originally from Dummit and Foote’s abstract algebra 3rd edition sec $1.3$ problem $11$. I tried to solve it and reduce the problem to the above. This is my attempt :

For the sake of contradiction let $$n(x-y)\equiv0\pmod m$$ Let $\\$ $n(x-y)=mk; k\in \mathbb Z^{+}$ $\\$ Since $x,y \lt m,x-y\text{ must be less than }m$.

Let $m=\prod_{i=1}^{\mathscr k}p_i^{j_i}$ and $k=\prod_{w=1}^{\mathscr l}q_{w}^{s_{w}}$ where $p_i,q_w$ are primes (not necessary distinct) and $j_i,s_w$ are positive integers.
We know that $n\in \mathbb Z^{+}$ from definition so $(x-y)|mk$
Since $x-y\lt m$ the proof can be divided into $3$ cases

First case
$(x-y)|k$ so it is equal to $\prod_{w=1}^{\mathscr s\le \mathscr l}q_w^{r_w}$ where $0\le r_w\le s_w$ and not all $r_w$ is $0$. So $n=m\cdot \prod_{w=\mathscr s+1}^{\mathscr l}q_w^{s_w-r_w}$, in this case the gcd of $n$ and $m$ is $\not=1$ Contradiction

Second case
$(x-y)|mk$ so it is equal to $\prod_{i=1}^{\mathscr z\le \mathscr k}\prod_{w=1}^{\mathscr c \le \mathscr l}p_{i}^{t_{i}}q_w^{v_w}$ where $0\le t_i \le j_i$(not all zero) and $0\le v_w \le s_w$(not all zero). Thus, $n=\prod_{i=\mathscr z+1}^{\mathscr k}\prod_{w=\mathscr c+1}^{\mathscr l}p_i^{j_i-t_i}q_w^{s_w-v_w}$. Since $n$ contains $\prod p_i$ the gcd of n and m $\not=1$

Third case
$(x-y)|m$. This case is similar to the first case so the proof is omitted.

For the converse to be true, we must prove that if $\gcd(n,m)\not=1$ then $n(x-y)\equiv0 \pmod m$
Let $gcd(n,m)=k\not=1$ then $n=kc$ and $m=kd$ where $\gcd (c,d)=1$
Since $x,y$ is in the residue class of $m$, one of the below is true.
$x-y\equiv0\pmod m$
$x-y\equiv1\pmod m$
$x-y\equiv2\pmod m$
$\vdots$
$x-y\equiv m-1\pmod m$
I am stuck here, I can’t prove that $n(x-y)\equiv0\pmod m$ for all these relations and I’m not even sure that I have taken the right approach. Also for the first part because $p_i$ and $q_w$ aren’t necessary distinct I think my prove doesn’t work. Could anyone please shed some light on this? A hint would be greatly appreciated.

Monai
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  • lcm stands for "least common multiple". The only way the least common multiple of nonzero integers is $1$ is if they are both equal to $1$. I suspect you mean gcd (greatest common divisor), which some confusingly denote lcf cor "least common factor", – Arturo Magidin Oct 08 '22 at 17:16
  • Also, as written your left hand side condition never holds, since you are allowed to take $x=y$. The statement it a mess. – Arturo Magidin Oct 08 '22 at 17:17
  • @ArturoMagidin I’m so sorry. I actually mean gcd as you said. I also forgot to add that $x\not=y$. – Monai Oct 08 '22 at 17:19
  • Hint: if $\gcd(n,m)=d$ and $m=dk$, then $n((k+1)-1)\equiv. 0\pmod{m}$. – Arturo Magidin Oct 08 '22 at 17:26
  • By the theorem in the linked dupe $f(x) = nx$ is $1$-$1$ on $\Bbb Z_m!\iff (n.m)=1\ \ $ – Bill Dubuque Oct 09 '22 at 04:04

1 Answers1

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$(\Leftarrow)$ Suppose $\gcd(n,m) = 1$. Then for all $x,y\in \{0,\cdots, m-1\},x\neq y, x-y\neq 0\mod m$ since $0<|x-y|<m$ and so $n(x-y)\not\equiv 0\mod m$. Otherwise since n is invertible modulo m, $x-y\equiv 0\mod m.$

$(\Rightarrow)$ Suppose $\gcd(n,m) = d > 1.$ Then $n(m/d) \equiv 0\mod m,$ so we can choose $x = 0, y = m/d > x.$

  • I don’t quite understand the last part how you can choose $x=0$ and $y=m/d$ when the proof ask for arbitrary $x$ and $y$ that are not equal. Could you please elaborate? – Monai Oct 09 '22 at 02:29
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Oct 09 '22 at 04:05