Question

3. Find all the solutions to the equation x^2-x=0 mod 12. Comment on your answer.

Answer 1

The solutions of the given equation are:

x=0,1,4 and 9

Explanation:

We are asked to find the solution of the equation:

`x^2-x=0\ \text{mod}\ 12`

i.e. we have to find the possible values of x such that the equation is true.

• If x=0

then

`x^2-x=0-0\\\\i.e.\\\\x^2-x=0`

Hence, x=0 is the solution of the equation.

• if x=1

then

`x^2=1\\\\Hence,\\\\x^2-x=1-1\\\\i.e.\\\\x^2-x=0`

Hence, x=1 is a solution.

• If x=2

then

`x^2=4`

i.e.

`x^2-x=4-2\\\\i.e.\\\\x^2-x=2\\eq 0`

Hence, x=2 is not a solution.

• If x=3

then

`x^2=9`

i.e.

`x^2-x=9-3\\\\i.e.\\\\x^2-x=6\\eq 0`

Hence, x=3 is not a solution.

• If x=4

then

`x^2=16=4\ \text{mod}\ 12`

i.e.

`x^2-x=4-4\\\\i.e.\\\\x^2-x=0`

Hence, x=4 is a solution to the equation.

• If x=5

then

`x^2=25=1\ \text{mod}\ 12`

i.e.

`x^2-x=1-4\\\\i.e.\\\\x^2-x=-3=9\ \text{mod}\ 12`

i.e.

`x^2-x=9\\eq 0`

Hence, x=5 is not a solution.

• If x=6

then

`x^2=36\\\\i.e.\\\\x^2=0\ \text{mod}\ 12\\\\i.e.\\\\x^2=0`

Hence,

`x^2-x=0-6\\\\i.e.\\\\x^2-x=-6=6 \text{mod}\ 12\\\\i.e.\\\\x^2-x=6\\eq 0`

Hence, x=6 is not a solution

• If x=7

then,

`x^2=49=1\ \text{mod}\ 12\\\\i.e.\\\\x^2=1`

Hence,

`x^2-x=1-7\\\\i.e.\\\\x^2-x=-6=6\ \text{mod}\ 12\\\\i.e.\\\\x^2-x=6\\eq 0`

Hence, x=7 is not a solution.

• If x=8

then,

`x^2=64=4\ \text{mod}\ 12`

i.e.

`x^2-x=4-8\\\\i.e.\\\\x^2-x=-4=8\ \text{mod}\ 12`

i.e.

`x^2-x=8\\eq 0`

Hence, x=8 is not a solution.

• If x=9

then,

`x^2=81=9\ \text{mod}\ 12`

i.e.

`x^2=9`

Hence,

`x^2-x=9-9\\\\i.e.\\\\x^2-x=0`

Hence, x=9 is a solution.

• If x=10

then,

`x^2=100=4\ \text{mod}\ 12`

i.e.

`x^2-x=4-10\\\\i.e.\\\\x^2-x=-6=6\ \text{mod}\ 12\\\\i.e.\\\\x^2-x=6\\eq 0`

Hence, x=10 is not a solution.

• If x=11

then,

`x^2=121=1\ \text{mod}\ 12`

i.e.

`x^2-x=1-11\\\\x^2-x=-10=2\ \text{mod}\ 12\\\\i.e.\\\\x^2-x=2\\eq 0`

Hence, x=11 is not a solution.