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.
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$$n(n+1)(n+2)(n+3)\equiv (-k-3)(-k-2)(-k-1)(-k)$$
$$\equiv (-1)^4(k+3)(k+2)(k+1)k\equiv k(k+1)(k+2)(k+3)\pmod{n+k+3}$$
It's true for all integers $n,k$ with no restrictions.
user236182
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This went over my head :( Could you elaborate? – András Szilveszter Oct 07 '15 at 19:47
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1I used modular arithmetic. Definition: $a\equiv b\pmod{n}$ means $n\mid a-b$. Then using this basic definition we can prove simple mod properties: see this post. – user236182 Oct 07 '15 at 19:52
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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.
Ángel Mario Gallegos
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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 $P(-k-3) = (-k-3)(-k-2)(-k-1)(-k) = k(k+1)(k+2)(k+3)$.
user133281
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$a-b\mid P(a)-P(b)$ is simple to prove: if $P(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_0$, then for all $k\in {0,1,2,\ldots, n}$:
$$c_ka^k-c_kb^k=c_k\left(a^k-b^k\right)=c_k(a-b)\left(a^{k-1}+a^{k-2}b+\cdots + b^{k-1}\right)$$
– user236182 Oct 08 '15 at 13:15