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

Question

asked Sep 17, 2020 222k views

1 Answer

Abuzar Amin · Answer 1 · 2020-09-22T07:49:32+0000

Answer:

$x=\pm√(77n+53)$

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
$x=\pm√(77(4)+53)$

$x=\pm√(308+53)$

$x=\pm√(361)$

$x=\pm 9$