1. ## 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.

2. Originally Posted by Rakesh
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, $(10/37) = (2/37)(5/37) = (-1)(1)=-1$.

3. Originally Posted by ThePerfectHacker
$(2/5) = 1$ 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, $(10/37) = (2/37)(5/37) = (-1)(-1)= 1$.

4. Originally Posted by Isomorphism
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

5. Originally Posted by Isomorphism
Shouldnt the above read like this?

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

Thus, $(10/37) = (2/37)(5/37) = (-1)(-1)= 1$.
Originally Posted by jtsab
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.