Proving there exists $k$ such that $p(n)p(n+1)=p(k)$

Question

Exact Question

Let $p(x)$ be monic quadratic polynomial over $\mathbb{Z}$. Show that for any integer $n$, there exists an integer $k$ such that $p(n)p(n+1)=p(k)$

Expanding the polynomial just creates a jumbled mess. Is there any intuitive way for this? Any hints appreciated

Answer 1

Let $p(x)=x^2+bx+c$ due to monicity (if this is even a word). Then $$\begin{align}p(n)p(n+1)&=(n^2+bn+c)((n+1)^2+b(n+1)+c)\\&=(n^2+bn+c)(n^2+(2+b)n+1+b+c)\\&=(n^2+bn+c)^2+(n^2+bn+c)(2n+1+b)\\&=(n^4+2bn^3+(b^2+2c)n^2+2bcn+c^2)+(n^3+(1+3b)n^2+(b^2+b+c)n+c+bc)\\&=n^4+(1+2b)n^3+(1+b^2+3b+2c)n^2+(b^2+b+2bc+c)n+(bc+c+c^2)\end{align}$$ Now if $p(n)p(n+1)=p(k)$ then $$n^4+(1+2b)n^3+(1+b^2+3b+2c)n^2+(b^2+b+2bc+c)n+(bc+c^2)=k^2+bk\tag{1}$$ for some $k$.

Firstly, to get $bc+c^2$, we must have that the coefficient independent of $n$ in $k$ is $c$ because then $k^2+bk$ 'ends' in $bc+c^2$.

Next, since $p(n)p(n+1)$ is also monic, so must $k$. Hence we currently have $$k=n^2+\text{something}\cdot n+c$$

Finally, substituting this into $(1)$ yields $$\boxed{k=n^2+(1+b)n+c}$$

Answer 2

If you put $p(n)=n^2+an+b$, then $p(n)p(n+1)=p(k)$ where $$k=n(n+1)+an+b.$$

Indeed, both $p(n)p(n+1)$ and $p(k)$ expand to $$ n^4 + (2a + 2)n^3 + (a^2 + 3a + 2b + 1)n^2 + (a^2 + (2b + 1)a + 2b)n + (ba + b^2 + b) $$

Answer 3

Noticing that we need to show $$(n-\alpha)(n-\beta)(n+1-\alpha)(n+1-\beta)=(k-\alpha)(k-\beta)$$ Also, $$(n-\alpha)(n+1-\beta)+\alpha=(n-\beta)(n+1-\alpha)+\beta$$ Putting both equals to $k$ we get our proof.