Legendre symbol

**ulysses123** · April 12th 2010, 02:02 AM

Calculate the legendre symbol using quadratic reciprocity $\frac{31}{43}$

I have calculated this and found it to be -1.
I am wondering if there is a way to check my answer. I tried to check on scientific notebook but that does not seem to have a function for the legendre symbol and i cannot find any online software for this.

**Deadstar** · April 12th 2010, 03:14 AM

Maple.

with(numtheory):

legendre(31,43);

I found the answer to be 1 though...

$\bigg{(} \frac{31}{43} \bigg{)} = \bigg{(} \frac{43}{31} \bigg{)} (-1)^{\tfrac{31-1}{2} \cdot \tfrac{43-1}{2}} = -\bigg{(} \frac{43}{31} \bigg{)}$

$= -\bigg{(} \frac{12}{31} \bigg{)}$

$= -\bigg{(} \frac{2^2}{31} \bigg{)} \cdot \bigg{(} \frac{3}{31} \bigg{)}$

$= -\bigg{(} \frac{3}{31} \bigg{)}$

$= -\bigg{(} \frac{31}{3} \bigg{)} (-1)^{\tfrac{31-1}{2} \cdot \tfrac{3-1}{2}}$

$= \bigg{(} \frac{31}{3} \bigg{)}$

$= \bigg{(} \frac{1}{3} \bigg{)} = 1$

**tonio** · April 12th 2010, 04:07 AM

Originally Posted by Deadstar

Maple.

with(numtheory):

legendre(31,43);

I found the answer to be 1 though...

$\bigg{(} \frac{31}{43} \bigg{)} = \bigg{(} \frac{43}{31} \bigg{)} (-1)^{\tfrac{31-1}{2} \cdot \tfrac{43-1}{2}} = -\bigg{(} \frac{43}{31} \bigg{)}$

$= -\bigg{(} \frac{12}{31} \bigg{)}$

$= -\bigg{(} \frac{2^2}{31} \bigg{)} \cdot \bigg{(} \frac{3}{31} \bigg{)}$

$= -\bigg{(} \frac{3}{31} \bigg{)}$

$= -\bigg{(} \frac{31}{3} \bigg{)} (-1)^{\tfrac{31-1}{2} \cdot \tfrac{3-1}{2}}$

$= \bigg{(} \frac{31}{3} \bigg{)}$

$= \bigg{(} \frac{1}{3} \bigg{)} = 1$

I got the same and in fact $31=17^2\!\!\!\pmod {43}$

Tonio

**ulysses123** · April 12th 2010, 05:57 PM

i had split 12/31
into 6 lots of 2/31 and then used the property that p is a prime of the form 8k-1 and thus 2/31 =1. Which works out the same except that i had forgotten the negative sighn from the first calculation of 31/43=-(43/31)

Thaks for the help though this method seems a bit easier.

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