Question

Solve the equation.

$x^2=\left[1\right]$in${\mathrm{\mathbb{Z}}}_8$

Short answer

$x=\left[0\right],\left[3\right],\left[4\right],\left[7\right]$is the solution of $x^2=\left[1\right],\left[3\right],\left[5\right],\left[7\right]$ . These are the solutions of $x^2=\left[1\right]$ in${\mathrm{\mathbb{Z}}}_8$ .

Step-by-step solution

Solve the equation

Consider the given equation

$x^2=\left[1\right]$in ${\mathrm{\mathbb{Z}}}_8$

$x$	$x^2$					Solution
$\left[0\right]$	$\left[0\right]$	$⊝$	$\left[0\right]$	$=\left[0\right]$	$=0$	No
$\left[1\right]$	$\left[1\right]$	$⊝$	$\left[1\right]$	$=\left[1\right]$	$=1$	Yes
$\left[2\right]$	$\left[2\right]$	$⊝$	$\left[2\right]$	$=\left[4\right]$	$=4$	No
$\left[3\right]$	$\left[3\right]$	$⊝$	$\left[3\right]$	$=\left[9\right]$	$=1$	Yes
$\left[4\right]$	$\left[4\right]$	$⊝$	$\left[4\right]$	$=\left[16\right]$	$=0$	No
$\left[5\right]$	$\left[5\right]$	$⊝$	$\left[5\right]$	$=\left[25\right]$	$=1$	Yes
$\left[6\right]$	$\left[6\right]$	$⊝$	$\left[6\right]$	$=\left[36\right]$	$=4$	No
$\left[7\right]$	$\left[7\right]$	$⊝$	$\left[7\right]$	$=\left[49\right]$	$=1$	YEs

Result

$x=\left[0\right],\left[3\right],\left[4\right],\left[7\right]$is the solution of $x^2=\left[1\right],\left[3\right],\left[5\right],\left[7\right]$ , and these are the solutions of $x^2=\left[1\right]$ in${\mathrm{\mathbb{Z}}}_8$ .