How to prove that $(k_{1}k_{2}...k_{n}) \equiv (z_{1}z_{2}...z_{n})\mod n$ if they are both complete system of residues modulo n.?

Question

Prove that if $(k_{1},k_{2},...,k_{n})$ and $(z_{1},z_{2},...,z_{n})$ are both complete system of residues $\bmod n$ then $$(k_{1}+k_{2}+...+k_{n}) \equiv (z_{1}+z_{2}+...+z_{n})\mod n$$ and $$(k_{1}k_{2}...k_{n}) \equiv (z_{1}z_{2}...z_{n})\mod n$$ can you give me some hints to complete this proof?

I know that proofs are not obvious but in this case

$(k_{1},k_{2},...,k_{n})\mod n$ and $(z_{1},z_{2},...,z_{n}) \mod n$

I probably need to apply some theorems in number theory for complete system of residues.

The basic step consists in proving that if $a\equiv a'$ and $b\equiv b'\mod n$, then $a+b\equiv a'+b'$ and $ab\equiv a'b'$, and to extend this result to any number of terms/factors. — Bernard, Dec 12 '20 at 10:29
Should I apply some specific theorems in number theory or not? — Charlie Van Basten Øydne, Dec 12 '20 at 10:34
Immediate consequence of the linked Congruence Sum and Product rules — Bill Dubuque, Dec 12 '20 at 10:42

score 1 · Answer 1 · answered Dec 12 '20 at 10:38

1

A complete residue system consists of a set of integers of the form:

$$\{0+a_0 n, 1+a_1 n, \dots, (n-1)+a_{n-1} n\}$$

where it is easy to see that addition and multiplication modulo $n$ give the same result.

answered Dec 12 '20 at 10:38

JMP

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How to prove that $(k_{1}k_{2}...k_{n}) \equiv (z_{1}z_{2}...z_{n})\mod n$ if they are both complete system of residues modulo n.?

1 Answers1