Quadratic Prime

Question

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.

Answer 1

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.

Answer 2

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

1. 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.
2. 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.
3. Show that this implies that $a=p$.
4. 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 $r_1r_2=s_1s_2+1$.
5. So $s_1=r_1+t$ and $s_2=r_2-t$ for some integer $t$. Check that $$1=r_1r_2-s_1s_2=t(r_1-r_2+t).$$
6. Show that the equation in step 5 is impossible unless $r_1=r_2$.