Quadratic Prime We had received some questions on Quadratic equations, But I wasnt able to do one. Here it goes:
Let $a,b$ be natural numbers $a>1$. Also, $p$ is a prime number. If $ax^2+bx+c=p$ for 2 distinct integral values of $x$. Then the number of integral roots of the equation $ax^2+bx+c=2p$ is ? Well I know the answer is $0$, But I am not able to get it properly. 

Once you are convinced that $a$ divides $b$,$c-p$ and $c-2p$ and that $a=p$, replace $b$ by $p\bar{b}$ and $c-p$ by $p\bar{c}$. Then the equations become (if we divide $p$ out) $x^2+\bar{b}x+\bar{c}$ and $x^2+\bar{b}x+\bar{c}-1$.
 Then look at the discriminants of both equations: they are $\sqrt{\bar{b}^2-4\bar{c}}$ and $\sqrt{\bar{b}^2-4\bar{c}+4}$. Since the discriminants should be integer, the expressions under the root have to be squares:
$\bar{b}^2-4\bar{c}=u^2$ and $\bar{b}^2-4\bar{c}+4 =v^2$ for some integers $u$ and $v$.
Then we have that $v^2-u^2=4$. A little calculation shows that $u=0$ and $v=\pm 2$. But $u$ cannot be zero since otherwise the discriminant of the first equation is also zero and both roots of the first equation coincide.
