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Congruence problem |
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| Oct23-05, 08:02 PM | #1 |
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Congruence problem
Hello I have the following question.
---- Solve the following congruences simultaneously: [tex]2x + 3y \equiv 2 mod 631[/tex] [tex]3x + 2y \equiv 3 mod 631[/tex] ---- I first tried adding and got [tex]5x + 5y \equiv 5 mod 631[/tex], but I was then stuck, so I tried the old multiplication, which looked worse as: [tex]6x^2 + 13xy + 6y^2 \equiv 6 mod 631[/tex] Any ideas? I am guessing that I need to go somewhere with the addition one, but I can't see where. The instructor had a hint of using the fact that 631 is prime, but I can't see anything from that. Thanks. Hmm, just got an idea, [tex]5(x + y) \equiv 5 mod 631[/tex], then find inverse of 5 and multiply it through to find x+y = something mod 631. I did this and got: [tex]x + y \equiv 5*79380 mod 631[/tex] which is [tex]x + y \equiv 1 mod 631[/tex] Now where can I go from here? Any ideas? |
| Oct24-05, 09:44 AM | #2 |
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Try solving them like regular simultaneous equations.
Eliminate x from the equations. You have a congruence relation for y. Now plug this back in and find a separate congruence relation for x. |
| Oct24-05, 03:31 PM | #3 |
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How would I do that without canceling the y's? In my mind I would have to multiply the top by 3 and bottom by 2, but when I added, I would get 0 = 0 mod 631.
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| Oct24-05, 06:42 PM | #4 |
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Congruence problem
No, you won't !
[tex]3x + 2y \equiv 3 ~~(mod~ 631)~~~~--(1)[/tex] [tex]2x + 3y \equiv 2 ~~(mod~ 631)~~~~--(2)[/tex] Let's do (1)*2 - (2)*3 ; this is legal since the modulo is the same for both congruences. [tex]6x + 4y \equiv 6 ~~(mod~ 631)~~~~--(3)[/tex] [tex]6x + 9y \equiv 6 ~~(mod~ 631)~~~~--(4)[/tex] Subtracting gives : [tex]5y \equiv 0~~ (mod~631) [/tex] What does this tell you about 'y' ? Plug that back in and figure out what 'x' should be. |
| Oct24-05, 06:50 PM | #5 |
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Wow, I am retarded
!!!! Thanks for pointing that out Gokul. Hmm, 3 times 3 is 9, and not 6, interesting! Thanks Gokul!
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| Oct24-05, 07:30 PM | #6 |
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[tex]2x + 3y \equiv 2 mod 631[/tex]
[tex]3x + 2y \equiv 3 mod 631[/tex] Did you notice that the right hand side is the same as the coefficients of x on the left? Hmm, suppose x= 1? |
| Oct25-05, 09:52 AM | #7 |
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