Let a,b be Natural numbers where a > 1. Also p is a prime number. If $ax^2 + bx + c = p$ for two distinct integer values of x, then prove that $ax^2 + bx + c \neq 2p$ for any integral values of x.
So this seems to be a pretty straightforward question and I got the final expression that is $p= a (x-y)(x-z)$ where y and z are roots of equations $ax^2 + bx + c - p$, so I got p has three factors while a prime number can only have 2 factors. But I couldn’t understand what would happen if $x-y = x-z = 1$. So essentially I need to try and prove that $y \neq z$. Thank you.
If $m,n;m\ne n$ are integer roots to $ax^2 + bx + c = p$ then $ax^2 + bx + c-p = a(x-m)(x-n)$
Now if $ax^2 + bx + c = 2p$ then $ax^2 + bx + c -p =a(x-m)(x-n)= p$. If $x$ is an integer and if $a > 1$ and $p$ is prime that means $a = p$ and $(x-m)(x-n) = 1$.
And $m,n$ are unequal integer then $x-m$ and $x-n$ are unequal integers. It's easy to verify that if $b,d$ are integers and $bd=1$ then $b=d= \pm 1$.