Quadratic reciprocity question

[math_grl]

Homework Statement

Show x^2 + (p+1)/4 \equiv 0 (\mod p) where p \equiv 3 (\mod 4) and p is prime is not solvable.

Homework Equations

Legendre's and Jacobi symbol, congruences

The Attempt at a Solution

Noticing that x^2 \equiv -(k+1) (\mod p) when p = 4k + 3 ?
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?

 

[Mark44]

Homework Statement

Show x^2 + (p+1)/4 \equiv 0 (\mod p) where p \equiv 3 (\mod 4).

Homework Equations

Legendre's and Jacobi symbol, congruences

The Attempt at a Solution

Noticing that x^2 \equiv -(k+1) (\mod p) when p = 4k + 3 ?
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?

I believe that your problem statement is incorrect, as it's not identically true for arbitrary value of x and p. Instead, I think it should instead ask for solutions of the equation.

Here's an example. Let p = 3.
Then the statement is
x² + (3 + 1)/4 \equiv 0 (mod 3).

Simplifying a bit gives
x² + 1 \equiv 0 (mod 3).

If x = 0, the above says that 1 \equiv 0 (mod 3) - not true.
If x = 1, we have 2 \equiv 0 (mod 3) - again not true.
If x = 2, we have 5 \equiv 0 (mod 3) - also not true.

Please clarify what it is that you need to do.

 

[math_grl]

I have edited the first post...i left out p is prime and we want to see show that the congruence is not solvable.