- #1
Ch1ronTL34
- 12
- 0
The question is:
Show that if p == 1 mod 4, then (a/p) = (-a/p).
(Note that == means congruent).
I know that if X^2==a mod p (p is a prime) is solvable then a is a quadratic residue of p.
For an example, I let p = 5 since 5==1 mod 4. Then, I let X = 2 and 4 just to check the equation. So:
2^2==-1 mod 5...a=-1 is quadratic residue.
4^2==1 mod 5...a=1 is quadratic residue.
I know that the legendre symbol (a/p) is 1 if a is a quadratic residue mod p and -1 if a is a quadratic non-residue.
From my example, (-1/5)=1 and (1/5)=1, so I have found an example that shows (a/p) = (-a/p) but I'm having trouble proving it in general.
Thanks!
Show that if p == 1 mod 4, then (a/p) = (-a/p).
(Note that == means congruent).
I know that if X^2==a mod p (p is a prime) is solvable then a is a quadratic residue of p.
For an example, I let p = 5 since 5==1 mod 4. Then, I let X = 2 and 4 just to check the equation. So:
2^2==-1 mod 5...a=-1 is quadratic residue.
4^2==1 mod 5...a=1 is quadratic residue.
I know that the legendre symbol (a/p) is 1 if a is a quadratic residue mod p and -1 if a is a quadratic non-residue.
From my example, (-1/5)=1 and (1/5)=1, so I have found an example that shows (a/p) = (-a/p) but I'm having trouble proving it in general.
Thanks!