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. 

A plan for proving that the latter equation cannot have integer roots, so assuming contrariwise. I left small gaps on purpose.


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*If a polynomial $Ax^2+Bx+C$ with integer coefficients has two
integer roots, then we necessarily have $A\mid B$ and $A\mid C$ as $-B/A$ is the sum of the roots and $C/A$ is their product.

*Show that if the latter equation has at least one integer root, then both of its roots are necessarily integers. From this point on the contrapositive assumption is thus that both equations have two integer roots.

*Show that this implies that $a=p$. 

*So $$ax^2+bx+(c-p)=p(x-r_1)(x-r_2)$$ and $$ax^2+bx+(c-2p)=p(x-s_1)(x-s_2)$$ for some integers $r_1,r_2,s_1,s_2$. Show that here $r_1+r_2=s_1+s_2$ and $r_1r_2=s_1s_2+1$.

*So $s_1=r_1+t$ and $s_2=r_2-t$ for some integer $t$. Check that
$$1=r_1r_2-s_1s_2=t(r_1-r_2+t).$$

*Show that the equation in step 5 is impossible unless $r_1=r_2$.

