1. ## Legendre Problem

Let p be an odd prime. Show that the Diophantine equation x^2+py+a=0; gcd(a, p)=1 has an integral solution if and only if (-a/p)=1.

2. ## Is correct this equation?

Hi,
I do not know if this helpful to you but:
Solving to y we have:
y=-x^2/p-a/p
Now if (-a/p) is not 1, there is still solution. Check for a=-15, p=11 (x=2, y=1).
If this is what you mean...

3. Originally Posted by gdmath
Hi,
I do not know if this helpful to you but:
Solving to y we have:
y=-x^2/p-a/p
Now if (-a/p) is not 1, there is still solution. Check for a=-15, p=11 (x=2, y=1).
If this is what you mean...
I am not sure what you are doing. The meaning of (-a/p) is not a fraction, it means this.

Originally Posted by cathwelch
Let p be an odd prime. Show that the Diophantine equation x^2+py+a=0; gcd(a, p)=1 has an integral solution if and only if (-a/p)=1.
There exists an $x$ and $y$ that solve this equation if and only if $x^2 + a = p(-y)$ if and only if $p|(x^2+a)$ if and only if $x^2\equiv -a(\bmod p)$, and this has a solution if and only if $(-a/p)=1$.

4. Sorry i just missunderstood the question.
I am not doing something suspicious, if this you mean.