1. ## modulo ?

hi there I having some troble with a few problems i was wondering if i could get some help?

x^2 equvilent to -1 mod 19

x^2 equvilent to -1 mod 19

4x equvilent to 6 mod 15

2. Originally Posted by bugal402
4x equvilent to 6 mod 15

I'll defer the first too to more able number theorists.

$\displaystyle 4x\equiv 6\text{ mod }15\implies 4x=6-15y\implies 4x+15y=6$ this is solvable since $\displaystyle 6=6(15,4)$. Now do you know how to solve Diophantine equations?

P.S. It might be easier to just guess :S

3. $\displaystyle (4,15)=1 \implies 4^{-1}$ exists modulo $\displaystyle 15$.

You can find $\displaystyle 4^{-1}$ by the Euclidean algorithm.

I omit details but $\displaystyle 4\cdot4\equiv1\bmod{15}\implies 4^{-1}\equiv4\bmod{15}$.

So $\displaystyle x\equiv 4^{-1}\cdot4x\equiv 4^{-1}\cdot6\equiv 4\cdot6 = 24\equiv 9 \bmod{15}$

4. Originally Posted by bugal402
hi there I having some troble with a few problems i was wondering if i could get some help?

x^2 equvilent to -1 mod 19

x^2 equvilent to -1 mod 19

You asked twice the same......anyway, the equation $\displaystyle x^2=-1\!\!\!\pmod p$ has a solution iff $\displaystyle p=1\!\!\!\pmod 4$ , so in your case...

Tonio

4x equvilent to 6 mod 15

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5. Originally Posted by Drexel28
I'll defer the first too to more able number theorists.

$\displaystyle 4x\equiv 6\text{ mod }15\implies 4x=6-15y\implies 4x+15y=6$ this is solvable since $\displaystyle 6=6(15,4)$. Now do you know how to solve Diophantine equations?

P.S. It might be easier to just guess :S
I know how to do Diophantine equations. thanks again

6. Originally Posted by tonio
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you are right I did the second one was supposed to be x^2 equvilent to -1 mod 17

7. Originally Posted by tonio
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So this one does not have a solution right?

8. Originally Posted by bugal402
So this one does not have a solution right?

The one for 19 hasn't , and the one for 17 has...and a pretty easy one, in fact.

Tonio