Modulo Arithmetic/Examples/Solutions to x^2 = x Modulo 6

Example of Modulo Arithmetic

The equation:

$x^2 = x \pmod 6$

has solutions:

\(\ds x\)

\(=\)

\(\ds 0\)

\(\ds x\)

\(=\)

\(\ds 1\)

\(\ds x\)

\(=\)

\(\ds 3\)

\(\ds x\)

\(=\)

\(\ds 4\)

Proof

The Cayley table for multiplication modulo $6$ can be presented as:

$\begin{array} {r|rrrrrr}

\times_6 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 \\ 2 & 0 & 2 & 4 & 0 & 2 & 4 \\ 3 & 0 & 3 & 0 & 3 & 0 & 3 \\ 4 & 0 & 4 & 2 & 0 & 4 & 2 \\ 5 & 0 & 5 & 4 & 3 & 2 & 1 \\ \end{array}$

The squares $x^2$ of each $x$ can be found on the main diagonal, where each element of $\Z_6$ can be inspected.

Checking each of these, we have:

\(\ds 0^2\)

\(=\)

\(\ds 0\)

\(\ds \)

\(\equiv\)

\(\ds 0\)

\(\ds \pmod 6\)

\(\ds 1^2\)

\(=\)

\(\ds 1\)

\(\ds \)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod 6\)

\(\ds 2^2\)

\(=\)

\(\ds 4\)

\(\ds \)

\(\equiv\)

\(\ds 4\)

\(\ds \pmod 6\)

\(\ds 3^2\)

\(=\)

\(\ds 9\)

\(\ds \)

\(\equiv\)

\(\ds 3\)

\(\ds \pmod 6\)

\(\ds 4^2\)

\(=\)

\(\ds 16\)

\(\ds \)

\(\equiv\)

\(\ds 4\)

\(\ds \pmod 6\)

\(\ds 5^2\)

\(=\)

\(\ds 25\)

\(\ds \)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod 6\)

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $7$