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#1

$Old$ November 2nd, 2009, 07:39 PM

jmedsy $jmedsy is offline$

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$Default$ find solution set for this linear congruence

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

Last edited by jmedsy; November 2nd, 2009 at 08:58 PM.

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#2

$Old$ November 2nd, 2009, 10:20 PM

alexmahone $alexmahone is offline$

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$x^2-2$ 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 $x=4 (mod\ 7)$ .

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$Old$ November 2nd, 2009, 10:41 PM

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thanks

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$Old$ November 2nd, 2009, 10:43 PM

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That's true, but its also true for x = 10. How can I solve for all possible x?

10 falls into the solution set $x=3 (mod\ 7)$ .

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#5

$Old$ November 2nd, 2009, 10:52 PM

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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?

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$Old$ November 2nd, 2009, 11:01 PM

alexmahone $alexmahone is offline$

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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.)

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$Old$ November 2nd, 2009, 11:03 PM

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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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