Respuesta :
Answer:
[tex]x=\pm\sqrt{77n+53}[/tex]
Step-by-step explanation:
Given : [tex]x^2\equiv 53\mod 77[/tex]
To find : All the square roots ?
Solution :
The primitive roots modulo is defined as
[tex]a\equiv b\mod c[/tex]
Where, a is reminder
b is dividend
c is divisor
Converting equivalent into equal,
[tex]a-b=nc[/tex]
Applying in [tex]x^2\equiv 53\mod 77[/tex],
[tex]x^2\equiv 53\mod 77[/tex]
[tex]x^2-53=77n[/tex]
[tex]x^2=77n+53[/tex]
[tex]x=\pm\sqrt{77n+53}[/tex]
We have to find the possible value in which the x appear to be integer.
The possible value of n is 4.
As [tex]x=\pm\sqrt{77(4)+53}[/tex]
[tex]x=\pm\sqrt{308+53}[/tex]
[tex]x=\pm\sqrt{361}[/tex]
[tex]x=\pm 9[/tex]
