Legendre symbol

Rakesh · #1 $Old$ May 6th, 2008, 04:18 PM

Determine the values of

47/37 and

3/43

stating any properties of the legendre symbol used.

is there just one method to solve all of these or are there different methods to solve depending on the numbers.

ThePerfectHacker · #2 $Old$ May 6th, 2008, 08:19 PM

Quote:

Originally Posted by Rakesh $View Post$

47/37

$(47/37) = (10/37)$ ---> Congruence
$=(2/37)(5/37)$ ---> Multiplicative

$(2/37) = -1$ because $37\equiv \pm 3(\bmod 8)$ .
$(5/37) = (37/5)$ ---> Quadradic Reciprocity
$(37/5) = (2/5)$ ---> Congruence
$(2/5) = 1$ because $5\equiv \pm 1(\bmod 8)$ .

Thus, $(10/37) = (2/37)(5/37) = (-1)(1)=-1$ .

Isomorphism · #3 $Old$ May 6th, 2008, 11:30 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

$(2/5) = 1$ because $5\equiv \pm 1(\bmod 8)$ .

Thus, $(10/37) = (2/37)(5/37) = (-1)(1)=-1$ .

Shouldnt the above read like this?

$(2/5) = -1$ because $5\equiv \pm 3(\bmod 8)$ .

Thus, $(10/37) = (2/37)(5/37) = (-1)(-1)= 1$ .

jtsab · #4 $Old$ May 7th, 2008, 01:30 AM

Quote:

Originally Posted by Isomorphism $View Post$

Shouldnt the above read like this?

$(2/5) = -1$ because $5\equiv \pm 3(\bmod 8)$ .

Thus, $(10/37) = (2/37)(5/37) = (-1)(-1)= 1$ .

That's what I thought, because $7 \equiv -1 \mod8$ and therefore not 5.

Also by Quadratic Reciprocity $\frac {2}{5} =-1$ as 2 is clearly not a square mod 5

ThePerfectHacker · #5 $Old$ May 7th, 2008, 10:36 AM

Quote:

Originally Posted by Isomorphism $View Post$

Shouldnt the above read like this?

$(2/5) = -1$ because $5\equiv \pm 3(\bmod 8)$ .

Thus, $(10/37) = (2/37)(5/37) = (-1)(-1)= 1$ .

Quote:

Originally Posted by jtsab $View Post$

That's what I thought, because $7 \equiv -1 \mod8$ and therefore not 5.

Also by Quadratic Reciprocity $\frac {2}{5} =-1$ as 2 is clearly not a square mod 5

Yes. It is exactly how Isomorphism said it should be.