Question

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

Answer 1

`x=\pm√(77n+53)`

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√(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√(77(4)+53)`

`x=\pm√(308+53)`

`x=\pm√(361)`

`x=\pm 9`