Legendre symbol

• May 6th 2008, 02:18 PM
Rakesh
Legendre symbol
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.
• May 6th 2008, 06:19 PM
ThePerfectHacker
Quote:

Originally Posted by Rakesh
47/37

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

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

Thus, $\displaystyle (10/37) = (2/37)(5/37) = (-1)(1)=-1$.
• May 6th 2008, 09:30 PM
Isomorphism
Quote:

Originally Posted by ThePerfectHacker
$\displaystyle (2/5) = 1$ because $\displaystyle 5\equiv \pm 1(\bmod 8)$.

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

Shouldnt the above read like this?

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

Thus, $\displaystyle (10/37) = (2/37)(5/37) = (-1)(-1)= 1$.
• May 6th 2008, 11:30 PM
jtsab
Quote:

Originally Posted by Isomorphism
Shouldnt the above read like this?

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

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

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

Also by Quadratic Reciprocity $\displaystyle \frac {2}{5} =-1$ as 2 is clearly not a square mod 5
• May 7th 2008, 08:36 AM
ThePerfectHacker
Quote:

Originally Posted by Isomorphism
Shouldnt the above read like this?

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

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

Quote:

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

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

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