$\displaystyle x^2-2$ is congruent to $\displaystyle 0(mod7)$
Last edited by jmedsy; Nov 2nd 2009 at 06:58 PM.
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Originally Posted by jmedsy $\displaystyle x^2-2$ is congruent to $\displaystyle 0(mod7)$ Note that $\displaystyle 3^2-2=7=0 (mod\ 7)$ and $\displaystyle 4^2-2=14=0 (mod\ 7)$. So $\displaystyle x=3 (mod\ 7)$ or $\displaystyle x=4 (mod\ 7)$.
thanks
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 $\displaystyle x=3 (mod\ 7)$.
Originally Posted by alexmahone 10 falls into the solution set $\displaystyle x=3 (mod\ 7)$. 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?
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.)
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?
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