Suppose $a,b,c$ are integers with $a>1$ , and $p$ is prime. Show that if $ax^2+bx+c$ is equal to $p$ for two distinct integral value of $x$,then $ax^2+ bx +c$ can not be equal to $2p$ for any integral value of $x$.
I'm stuck with this question. I approached it by considering the relationship between the coefficients and roots but I'm still stuck. Help me out and please give me the intuition behind your approach to this problem. I would be happy to learn.
Suppose that $ax^2 + bx + c = p$ has two distinct integral solutions $x=m,n$. So we factorise $ax^2+bx+c-p$ as $a(x-m)(x-n)$ (we know that $x-m$ and $x-n$ are factors by the factor theorem. Now we wish to show that $a(x-m)(x-n) = p$ has no solutions. Assume for contradiction there exists an integral solution. Note that since $p$ is prime, two of these factors must both be $1$ and the third must be $p$ (or negatives in two of these factors). But $a>1$, so $a \neq \pm 1$ and we have $a=p$, and $x-m=x-n=1$ or $-1$ for some $x$. But clearly $x-m = x-n$ is a contradiction because $m$ and $n$ are distinct.
The intuition here was to use the factor theorem, and try and derive some contradiction from the fact that $p$ is prime; it made sense to show that $p$ would end up being written as the product of at least two numbers and yield a contradiction.