# Thread: find solution set for this linear congruence

1. ## find solution set for this linear congruence

$x^2-2$ is congruent to $0(mod7)$

2. Originally Posted by jmedsy
$x^2-2$ is congruent to $0(mod7)$
Note that $3^2-2=7=0 (mod\ 7)$ and $4^2-2=14=0 (mod\ 7)$.
So $x=3 (mod\ 7)$ or $x=4 (mod\ 7)$.

3. thanks

4. Originally Posted by jmedsy
That's true, but its also true for x = 10. How can I solve for all possible x?
10 falls into the solution set $x=3 (mod\ 7)$.

5. Originally Posted by alexmahone
10 falls into the solution set $x=3 (mod\ 7)$.
Once you had determined that x=3 or x=4 satisfied the congruency, did you continue to check congruency for greater x? How did you know when to stop?

6. Originally Posted by jmedsy
Once you had determined that x=3 or x=4 satisfied the congruency, did you continue to check congruency for greater x? How did you know when to stop?
You need to check congruency only for x ranging from 0 to 6. (If the modulus was n, you should have checked for x ranging from 0 to n-1.)

7. right, that makes sense. Is there any correlation between the degree of that polynomial and the number of different modular congruences between 0 and 6?