Let $p \in \mathbb Z [x]$ be a monic quadratic polynomial. Show that for $n \in \mathbb Z$ there exists $k \in \mathbb Z$ such that $$p(n)p(n+1)=p(k)$$
I came across this problem in an olympiad book. I have tried this question by many different approaches like simply putting $n , (n+1)$ into the quadratic equation and and try to factorise or do something to get the result but it doesn't help. Secondly, I know that
$$p(x)-p(n)p(n+1)=0$$
has a root equal to $k$ and then there is only one thing left to do that is to prove k integer which we can prove by proving discriminant to be a perfect square and other coefficients as integer but the problem arises how to prove discriminant to be a perfect square. Please tell if either of these approaches will take us to the prove or if not please tell the correct steps of proving
I am very confused with problem (not a homework problem). Please help to prove it with proper steps and methods.
Write $p(x)=x^2+ax+b$.
Solution #$1$:
The intuition is that it's probably an algebraic identity.
Since $p(x)p(x+1)$ expands to a monic polynomial of degree $4$, it makes sense to look for an identity of the form $$p(x)p(x+1)=p(x^2+cx+d)$$ Expanding and equating coefficients, we get $c=a+1$ and $d=b$, which yields the verifiable identity $$p(x)p(x+1)=p(x^2+(a+1)x+b)$$
Solution #$2$:
As an alternative method, we can follow through with your approach . . .
Fix an integer $n$.
Our goal is to show that the quadratic equation $$p(x)=p(n)p(n+1)$$ has an integer solution for $x$.
Equivalently, we want to show that the discriminant $D$ is a perfect square.
Computing $D$, we get \begin{align*} D&= 4n^4\\[2pt] &{\phantom{=}}\;+(8+8a)n^3\\[2pt] &{\phantom{=}}\;+(4a^2+12a+8b+4)n^2\\[2pt] &{\phantom{=}}\;+(4a^2+8ab+4a+8b)n\\[2pt] &{\phantom{=}}\;+(a^2+4ab+4b^2)\\[6pt] &=\bigl(2n^2+(2a+2)n+(a+2b)\bigr)^2 \end{align*}