1. Legendre Symbol (5/p)

Hi, I have this problem.

Let p be an odd prime. Show that the legendre symbol (5/p)= 1 if and only if $p\equiv{-1}(mod 5)$.

I am really struggling to understand this legendre symbol stuff so if anyone could give me a clue I would be very grateful.

Thanks,
Katy

2. Originally Posted by harkapobi
Hi, I have this problem.

Let p be an odd prime. Show that the legendre symbol (5/p)= 1 if and only if $p\equiv{-1}(mod 5)$.

I am really struggling to understand this legendre symbol stuff so if anyone could give me a clue I would be very grateful.

Thanks,

Katy

By quadratic reciprocity, $\binom {5}{p}=\binom{p}{5}$ , since $5=1\!\!\!\pmod 4$, and $\binom{p}{5}=1\Longleftrightarrow p=\pm 1\!\!\!\pmod 5$ ...and thus the claim is false, and indeed:we've, for example, $\binom{5}{11}=1$ since $4^2=5\!\!\!\pmod{11}$, but nevertheless $11\ne -1\!\!\!\pmod 5$ .

Tonio

3. The Legendre symbol (5/p) =1 means that 5 is a quadratic residue mod p. The number 4^2 = 16 = 5 (11) shows that 5 is a quadratic residue mod 11, however 11 = 1(5), not -1.

4. Oh, yes your absolutely right, the question said $p\equiv{\pm1}(mod 5)$

So because and $5\equiv1(mod 4)$
$\binom {5}{p}=1\equiv{p^2}(mod 5)$

$\Rightarrow {p^2}\equiv1(mod 5) \Rightarrow p\equiv{\pm1}(mod 5)$

Is this right? How do you know that $\binom {p}{5} = 1$?

5. Originally Posted by harkapobi
Oh, yes your absolutely right, the question said $p\equiv{\pm1}(mod 5)$

So because and $5\equiv1(mod 4)$
$\binom {5}{p}=1\equiv{p^2}(mod 5)$

$\Rightarrow {p^2}\equiv1(mod 5) \Rightarrow p\equiv{\pm1}(mod 5)$

What is this?? What does $p^2$ have to do in all this?

Is this right? How do you know that $\binom {p}{5} = 1$?

Because the only quadratic residues modulo 5 are 0, 1, -1. As p is a prime number it can't be 0 mod 5, so...

Tonio

6. The $p^2$ comes from:

$\binom {n}{p}\equiv{n^{0.5(p-1)}}(mod p)$
is this not right?

Then:
$\binom {p}{5}\equiv{p^{0.5(5-1)}} = {p^2}(mod p)$

If this is not right, I dont understand how you get $p\equiv{\pm1}(mod 5)$

7. You've already done the hard work. Once you know that
$
\binom {5}{p} = \binom {p}{5}
$

you just have to look at the quadratic residues mod 5. Now any number, not just a prime p, is equivalent to 1,2,3 or 4 mod 5 (we ignore numbers divisible by 5). So we test them 1 by 1.

$
1^2 = 1 \equiv 1 mod 5
$

$
2^2 = 4 \equiv -1 mod 5
$

$
3^2 = 9 \equiv 4 \equiv -1 mod 5
$

$
4^2 = 16 \equiv 1 mod 5
$

So we've tested all possible numbers and found that 1 and -1 are the only quadratic residues, and that 2 and 3 are not quadratic residues for any number. It's as simple as that.

8. Originally Posted by harkapobi
The $p^2$ comes from:

$\binom {n}{p}\equiv{n^{0.5(p-1)}}(mod p)$
is this not right?

Then:
$\binom {p}{5}\equiv{p^{0.5(5-1)}} = {p^2}\pmod p$

But then here it is $\binom {p}{5}\equiv p^2\!\!\!\!\pmod 5$ , ain't it?

Tonio

If this is not right, I dont understand how you get $p\equiv{\pm1}(mod 5)$
.

9. Yes your right I meant to write mod 5. Sorry.

So what do you do when you cannot know if $\binom {p}{q} = \binom {q}{p}$ or $\binom {p}{q} = -\binom {q}{p}$?

For example $\binom {7}{p}$. You dont know if $\binom {7}{p} \binom {p}{7}$ is +1 or -1.

Do you just look at both cases? Because I am suppose to prove that $\binom {7}{p} = 1\Longleftrightarrow p\equiv{\pm1}, {\pm3} or {\pm9}(mod 28)$

I dont understand where the (mod 28) comes from here. I really dont understand this stuff very well.

Thank you for all your help
Katy

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5 above p legendre symbol

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