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let $p$ be a prime and let $a$ be an integer not divisible by $p$, that is $(a,p)=1$. Then $\{a,2a,3a,...,pa\}$ is a complete residue system modulo $p$.

Since $p\nmid a$, $p\nmid\{a,2a,3a,...,(p-1)a\}$ Since $p|p$, $p|\{pa\}$ which is exactly one element in the set and is therefore a complete residue system.

Wondering if this is an acceptable proof to this theorem.

Jordan L
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2 Answers2

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$$k_1a \equiv k_2a \pmod p \iff p \ | \ (k_1 - k_2)a \iff p \ | \ (k_1 - k_2)$$

Since in our case $k_1, k_2 \in A = \{1, 2, ..., p\}$ we have that $\left| { k_1 - k_2} \right| \lt p $ and $p$ will not divide $(k_1 - k_2)$ if both elements are from $A$. So for any two elements $r, q \in A \;\; rp \not \equiv qp \pmod p $. Which says that every element in $\{a,2a,3a,...,pa\}$ is not congruent to each other modulo $p$. Then each of them must be distinct modulo $p$. Since we have $p$ elements in the set it forms a complete set of residues modulo $p$.

Ishfaaq
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${\rm mod}\ p\!:\ f(x) = x^p\!-x \equiv(x\!-\!1)(x\!-\!2)\cdots(x\!-\!p)\, $ by $\,f\,$ has roots $\,1,\ldots,p,\,$ by little Fermat.

Thus $\ \ (a^{-1}x)^p\!-a^{-1}x \equiv (a^{-1}x\!-\!1)(a^{-1}x\!-\!2)\cdots(a^{-1}x\!-\!p)\,\ $ by $\,\ x\mapsto a^{-1}x$

Thus we infer $\ \ x^p-x\equiv (x\!-\!a)(x\!-\!2a)\cdots(x\!-\!pa)\ $ by multiply LHS by $\,a,\,$ RHS by $\ a^p\equiv a$

Therefore $ \{1,2,\ldots,p\}\equiv \{a,2a,\ldots,pa\},\,$ both being all roots of $\,f(x)\,$ in the field $\,\Bbb Z/p$.

Bill Dubuque
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  • Invoking Fermat's little theorem is overkill... and in fact this is a lemma often proved in the course of proving Fermat's. – Álvaro Lozano-Robledo Mar 04 '14 at 01:27
  • @Álvaro It's meant to be a proof to complement the prior-posted standard-proof. There are many ways to prove little Fermat - just as there are may ways to prove this. The more ways one knows, the more insight one gains. – Bill Dubuque Mar 04 '14 at 01:30