Thread: modulo ?

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

thanks for your help

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

thanks for your help
I'll defer the first too to more able number theorists.

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

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

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

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

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

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

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 $x^2=-1\!\!\!\pmod p$ has a solution iff $p=1\!\!\!\pmod 4$ , so in your case...

Tonio

4x equvilent to 6 mod 15

thanks for your help
.

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

$4x\equiv 6\text{ mod }15\implies 4x=6-15y\implies 4x+15y=6$ this is solvable since $6=6(15,4)$. 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
.
you are right I did the second one was supposed to be x^2 equvilent to -1 mod 17

sorry about that

7. Originally Posted by tonio
.
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