Proof that $n+k+3$ divides $n(n+1)(n+2)(n+3) - k(k+1)(k+2)(k+3)$.

Question

I'm looking for proof that $$ (n+k+3) \mid n(n+1)(n+2)(n+3) - k(k+1)(k+2)(k+3)\\ n,k \in \mathbb N^*, n>k $$ I tried using induction, but i'm not sure how it would work with 2 parameters.

Answer 1

Since $$n+3=n+k+3-k\qquad \text{and}\qquad k+3=n+k+3-n$$ we have \begin{align} n(n+1)(n+2)(n+3)&=n(n+1)(n+2)(n+k+3)-n(n+1)(n+2)k\\ k(k+1)(k+2)(k+3)&=k(k+1)(k+2)(n+k+3)-k(k+1)(k+2)n \end{align} Then it will be sufficient to show that $n+k+3$ divides $k(k+1)(k+2)n-n(n+1)(n+2)k$

But \begin{align} k(k+1)(k+2)n-n(n+1)(n+2)k&=nk(k^2+3k-n^2-3n)\\ &=nk\left[-(n-k)(n+k)-3(n-k)\right]\\ &=-nk(n-k)(n+k+3) \end{align}

Which completes the proof.

Answer 2

$$n(n+1)(n+2)(n+3)\equiv (-k-3)(-k-2)(-k-1)(-k)$$

$$\equiv (-1)^4(k+3)(k+2)(k+1)k\equiv k(k+1)(k+2)(k+3)\pmod{n+k+3}$$

It's true for all integers $n,k$ with no restrictions.

Answer 3

If $P$ is a polynomial with integer coefficients, then $$a-b \mid P(a)-P(b)$$ for integers $a$ and $b$. The desired conclusion follows by applying this with $a=n$, $b=-k-3$ and $$P(x) = x(x+1)(x+2)(x+3).$$ Indeed, we have $P(n)=n(n+1)(n+2)(n+3)$ and $P(-k-3) = (-k-3)(-k-2)(-k-1)(-k) = k(k+1)(k+2)(k+3)$.