1. ## LQR problem

Can someone check my argument on this problem? If p is prime and $p\equiv 1 ( mod 4)$ and $m^2+n^2=p$ and $m>0$ odd. Then $(\frac{m}{p})=1$
By LQR,
$(\frac{m}{p})=(\frac{p}{m})=(\frac{m^2+n^2}{m})=(\ frac{n^2}{m})$. Here, do I need to check the cases when m is prime and when m is composite to figure out if $(\frac{n^2}{m})=1$?
In class, I think we have a property that $(\frac{n^2}{p})=1$ with p prime.

2. Originally Posted by namelessguy
$(\frac{m}{p})=(\frac{p}{m})$?
I'm not sure I get that line

3. Originally Posted by SimonM
I'm not sure I get that line
He's using the Legendre symbol, if that's what you don't get.

4. Originally Posted by alexmahone
He's using the Legendre symbol, if that's what you don't get.
I get that, I just don't quite get what he has done which allows that, it looks like quadratic reciprocity, but not quite

5. Originally Posted by SimonM
I get that, I just don't quite get what he has done which allows that, it looks like quadratic reciprocity, but not quite
Hi SimonM
$(\frac{m}{p})=(\frac{p}{m})$ since $p\equiv ( 1 mod 4 )$

6. Originally Posted by alexmahone
He's using the Legendre symbol, if that's what you don't get.
Originally Posted by SimonM
I get that, I just don't quite get what he has done which allows that, it looks like quadratic reciprocity, but not quite
This is all okay. Because namelessguy is using Quadradic Reciprocity generalized to Jacobi symbols.

7. Originally Posted by namelessguy
Here, do I need to check the cases when m is prime and when m is composite to figure out if $(\frac{n^2}{m})=1$?
In class, I think we have a property that $(\frac{n^2}{p})=1$ with p prime.
To prove that, all you need to show is that (n^2,m)=1. Can you do this?

Assume the opposite

8. Originally Posted by SimonM
To prove that, all you need to show is that (n^2,m)=1. Can you do this?

Assume the opposite
I only need to know that $(n,m)=1$, which is obviously true from the hypothesis, we know that $n, m, p$ are relatively prime. Then I use a theorem for Jacobi symbol to conclude that $(\frac{n^2}{m})=1$