Congruences and Diophantine equations

**Willie_Trombone** · February 8th 2009, 07:59 PM

How do I use congruences to solve Diophantine equations?
For example 12x+25y=331.
I know this can be expressed as 12x=331(mod 25) but I don't know how to solve without just going through the numbers 1-25 to see which one works.

also how can I find the number of solutions for x^3=x^2(mod 50)?

**ThePerfectHacker** · February 8th 2009, 08:05 PM

Originally Posted by Willie_Trombone

How do I use congruences to solve Diophantine equations?
For example 12x+25y=331.
I know this can be expressed as 12x=331(mod 25) but I don't know how to solve without just going through the numbers 1-25 to see which one works.

also how can I find the number of solutions for x^3=x^2(mod 50)?

Notice, $12(-2) + 25(1) = 1 \implies 12(-662) + 25( 331) = 331$ .
Therefore, all solutions are given by $x=-662 + 25t \text{ and }y=331 - 12t$ , $t\in \mathbb{Z}$ .

also how can I find the number of solutions for x^3=x^2(mod 50)?

This can be written as $x^2(x-1)\equiv 0(\bmod 50)$ .
Therefore, $x^2(x-1)\equiv 0 (\bmod 2) \text{ and }x^2(x-1)\equiv 0(\bmod 25)$ .
Can you solve?

**Willie_Trombone** · February 8th 2009, 08:53 PM

haha, sorry I'm still not sure what to do at that point in the second problem. I just really can't seem to get the hang of this material.

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