Let $a,b \in N$ and $p$ be a prime . If $ax^2+bx+c=p$ has two distinct integral values of $x$ .Show that $ax^2+bx+c = 2p$ has no integral value.

Question

This is the way I approached the problem

Suppose that $b$ is even.

Then for the equation to have a integral root, $b^2-4a(c-p)$ must be a perfect square of an even number,say $4n^2$.(Because the numerator should be divisible by $2$.)

Also suppose equation $2$ has an integral solutionn, Then $b^2-4a(c-2p)=4m^2$ .

Thus $4(n-m)(n+m)=4ap$.

Now I should somehow prove that $p$ cannot be a prime for this to happen . Am I right ?

Answer 1

The result doesn't seem to be true. $x^2+x+2=2$ and $x^2+x+2=4$ both have two integer solutions.