Let $d=2\bmod4$ and $D=d^3-1$. We will show that the elliptic curve $E_D: y^2=x^3+D$ has no integer solutions $x,y$.
At first we show that $x$ is odd which is simple. Then we have the following argument:
$y^2+1=x^3+d^3=(x+d)(x^2-dx+d^2)$. Since $x$ is odd we have $x^2=1\bmod4$ and $dx=2\bmod4$ then $x^2-dx+d^2=3\bmod4$.
Then I cannot understand the following argument:
Discriminant of $x^2-dx+d^2$ is negative so it is divisible by a prime $p=3\bmod4$.
I understand the final argument: so $y^2+1$ is divisible by $p$ then $y^2=-1\bmod4$ but $-1$ is not a square. Contradiction.
Can someone explain me the argument above? Thanks for your help.
This example appears in Silverman's book "Arithmetic of Elliptic Curves".
The argument:
You already know that $x^2 - dx + d^2\equiv 3 \mod 4$. The discriminant argument tells us that $x^2 - dx + d^2$ must be positive. Thus it is a positive odd integer.
It follows that at least one of its prime factor must be $\equiv 3 \mod 4$: otherwise, all of its prime factors would be $\equiv 1\mod 4$, and hence it would also be $\equiv 1\mod 4$, a contradiction.