Given is prime number $p$ and natural number $a$ which is relatively prime to $p$. Prove that no numbers from the set $B={0a,1a,2a,...,(p-1)a}$ give the same value after divison by $p$.
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1if $p$ divides $(n-m)a$ and $p\nmid a$ then $p$ divides $(n-m)$ – J. W. Tanner Nov 18 '19 at 17:28
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i.e. $,x\mapsto ax,$ is injective ($1$-$1$) on $,\Bbb Z_p,,$ which follows by the Theorem in the linked dupe. – Bill Dubuque Nov 18 '19 at 17:54
3 Answers
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Suppose there are such numbers: $ma$ and $na$ ($n>m$), this would mean, that $p|a(n-m)$ since $p$ and $a$ are relative prime, this lead to $p|(n-m)$. But that is impossible, since $0<n-m<p$. Hence no such numbers exist and every element of the set gives different value$\pmod{p}$.
Andronicus
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If $p$ divides $r_1a-r_2a=(r_1-r_2)$ where $p-1\ge r_1>r_2\ge1\ \ (1)$
We need $p(r_1-r_2)$
But by $(1),$ $$0<r_1-r_2<p-1$$ which is not divisible by $p$
lab bhattacharjee
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What if they did?
Suppose that $ak \equiv aj \equiv m\pmod p$ for some $0 \le m < p$. What would that mean.
It would mean that $ak = m + p*w$ for some integer $w$ and then $aj = m + p*v$ for some $v$.
If we subtract those then $ak - aj = p*w -p*v$ so $a(k-j) = p(w-v)$
So $p|a(k-j)$.
Can you finish?
fleablood
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