How to prove that $f(x)=x^3 \pmod{pq}$ is bijective for any non negative integer $x

Question

I am reading a book on cryptographic programming and I found an example without proof.

How to prove that $f(x)=x^3 \pmod{pq}$ is bijective for any non negative integer $x<pq$ where 3 is not a factor of $p-1$ and $q-1$?

I did some experiments with Mathematica and I noticed the claim is true.

p = 11;
q = 17;
A = Range[0, p q - 1];
B = A^3 // Mod[#, p q] & // Sort;
A == B

I have no idea how to start proving this.

Note: $p$ and $q$ are two distinct large prime numbers.

The screenshot of the example:

There must be stated conditions on what $p$ and $q$ can be, because this claim is not true for $p = q = 2$. There is no nonnegative integer $x$ whose cube is congruent to 2 mod 4. — user1090793, Sep 10 '22 at 06:02
Dupe of Uniqueness of $k$th root mod $m$ if $(k, \phi(m)) = 1$. since $(3,\phi(pq)) = 1$ so by here $,x\to x^3,$ is surjective(onto) hence bijective ($1$-$1$) by pigeonhole. — Bill Dubuque, Sep 10 '22 at 08:18

score 3 · Accepted Answer · answered Sep 10 '22 at 06:15

Well you might begin by observing that if $x^3\equiv y^3$ then $x^3-y^3=(x-y)(x^2+xy+y^2)\equiv 0$

Then $pq$ cannot be a factor of $x-y$ because $x$ and $y$ are too small by hypothesis.

So this means that if $x^3\equiv y^3$ then either $p$ or $q$ (or both) must be a factor of $x^2+xy+y^2$.

You haven't said what you already know, but there is an extra condition given on $p$ and $q$ and there are standard mathematical techniques for dealing with a quadratic (taking out the factor $x-y$ has reduced the degree). So you may be able to take it from there.

score 2 · Answer 2 · edited Sep 10 '22 at 06:48

2

It is relatively elementary to prove that $f$ is bijective on the four following (invariant) subsets, which form a partition of $\mathbb Z/(pq\mathbb Z):$ $$\{0\},\{(pk)\bmod{pq}:q\nmid k\},\{(qk)\bmod{pq}:p\nmid k\}, \{k\bmod{pq}:p\nmid k,q\nmid k\}.$$

edited Sep 10 '22 at 06:48

Ralph Clausen

2,371

answered Sep 10 '22 at 06:29

Anne Bauval

49,005

How to prove that $f(x)=x^3 \pmod{pq}$ is bijective for any non negative integer $x

2 Answers2