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Show that if $m$ is odd then $\frac{(-m+1)}{2}, \frac{(-m+3)}{2}, \cdots, \frac{(m-3)}{2}, \frac{(m-1)}{2}$ is a complete residue system modulo m.

I understand the definition of a complete residue system but is having a hard time incorporating the odd part of this statement.

Bill Dubuque
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Abbi
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    Suggestion: take $m$ to be an even number (just pick one), calculate the listed elements and see why it doesn't work. – David Feb 17 '15 at 04:17

2 Answers2

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If $r_0, \ldots, r_{m-1}$ is a complete residue system modulo $m$ and $k$ is an integer, then $r_0-k, \ldots, r_{m-1}-k$ is a complete residue system modulo $m$ as well.

Note that $k=\frac{m-1}{2}$ is an integer as $m$ is an odd number. Since $0, 1, \ldots, m-1$ is obviously a complete residue system modulo $m$ we conclude that $$ \frac{(-m+1)}{2}=0-\frac{m-1}{2},\frac{(-m+3)}{2}=1-\frac{m-1}{2}, \ldots, \frac{(m-1)}{2}=m-1-\frac{m-1}{2}$$ is a complete residue system modulo $m$, as well.

Janko Bracic
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Let $\,m = 2n\!+\!1\,$ so $\,(m\!-\!1)/2 = n,\,$ Your sequence is $\,-n,-(n\!-\!1),\ldots,-1,0,1,\ldots,n\!-\!1,n,\,$ which is a sequence of $\,m = 2n+1\,$ consecutive integers. It is just the standard complete residue system (CRS) shifted left by $\,n,\,$ i.e. mapped by $\,x\mapsto x-n,\,$ which is a bijection mod $\,m.\,$

There is no standard name for this CRS, but you will sometimes see it called one of the following: least magnitude, numerically least, balanced, symmetric, signed.

Generally, by the Pigeonhole Principle, any set of $\,m\,$ integers that are pairwise incongruent $\!\bmod m\,$ form a CRS, for example any $\,m\,$ consecutive integers in an arithmetic progression which have step-size $\delta$ coptime to $\,m\,$ (OP is consecutive case where $\delta =1)$.

Remark $ $ Least magnitude residues are sometimes preferred since they may introduce simplifications, e.g. the smaller numbers may simplify calculations, e.g. using $(\pm1)^n$ vs. $9^n$ or $11^n$ in casting out nines and elevens, and as here when computing $\!\bmod\!$ on Gaussian integers $\,a+bi$ by rationalizing the modulus, where least-magnitude yields a smaller norm remainder (which enables descent in the Gaussian integer Euclidean algorithm). Further, the balanced symmetry may help exploit innate reflection (negation) symmetry, e.g. in this additive form of Wilson's theorem (sum of all elements $\!\bmod n$).

See also here and here for conventions used when extending the (Euclidean) division with remainder algorithm from $\Bbb N$ to $\Bbb Z$ (negative moduli and remainders).

Bill Dubuque
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