- #1
math_grl
- 49
- 0
Homework Statement
Show [tex]x^2 + (p+1)/4 \equiv 0 (\mod p)[/tex] where [tex]p \equiv 3 (\mod 4)[/tex] and p is prime is not solvable.
Homework Equations
Legendre's and Jacobi symbol, congruences
The Attempt at a Solution
Noticing that [tex]x^2 \equiv -(k+1) (\mod p)[/tex] when [tex]p = 4k + 3[/tex] ?
Now (-1/p)(k+1/p) should tell use whether this has a solution.
But (-1/p)=-1. How do you get (k+1/p)? Am I even on the right track?
Does (k+1/p) = (-1) (p/k+1)= (-1) (-1/k+1)? plus who's to say that k+1 is not factorable?
Last edited: