Question

Find all the square roots of x^2 = 53 (mod 77) by hand. 2 marks

Answer 1

Answer:

$x=\pm\sqrt{77n+53}$

Step-by-step explanation:

Given : $x^2\equiv 53\mod 77$

To find : All the square roots ?

Solution :

The primitive roots modulo is defined as

$a\equiv b\mod c$

Where, a is reminder

b is dividend

c is divisor  

Converting equivalent into equal,

$a-b=nc$

Applying in $x^2\equiv 53\mod 77$,

$x^2\equiv 53\mod 77$

$x^2-53=77n$

$x^2=77n+53$

$x=\pm\sqrt{77n+53}$

We have to find the possible value in which the x appear to be integer.

The possible value of n is 4.

As $x=\pm\sqrt{77(4)+53}$

$x=\pm\sqrt{308+53}$

$x=\pm\sqrt{361}$

$x=\pm 9$