Formula for monic quadratic polynomials over $\mathbb{Z}$

Question

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

Answer 1

Hint: $\;\;\;\;k = p(n)+n\;\;\;\;$

Answer 2

Let P(x)= x^2 + bx + c Therefore, P(n)= n^2 + bn + c, for an arbitrary integer n

P(n+1) = (n+1)^2 + b(n+1) + c = n^2 + 2n + 1 + bn + b + c = P(n) + 2n + b + 1

Therefore, P(n)P(n+1) = P(n)*{P(n) + 2n + b + 1} = {P(n)}^2 + 2nP(n) + bP(n) + P(n) = {P(n)}^2 + 2nP(n) + n^2 + bn + bP(n) +c = (P(n) + n}^2 + b{P(n) + n} + c = k^2 + bk + c , where k = P(n) + n, (i.e) an integer, = P(k)