# Proving by strong induction that $\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)…(n+d-1)$

I'm trying to prove by induction the following statement without success:
$$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1)$$

For the base case: $n = 2$, $d = 2$
$2\mid 2(2+1)$ which is true.

Now, the confusion begins! I assume I would need to use the second induction principle to proof this because $P(n)$ and $P(n+1)$ are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me.

I tried the following:
- Trying to prove by simple induction. I did not go very far.
- Trying to split my induction step in 2 parts: If $d\mid n$, I'm done.

If $d$ does not divide $n$, then I would need to do a second proof and this is where I'm blocked.

Anyone could tell me what's wrong in the approach I take to solve this problem?

Any help would be appreciated!

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you only need one induction, on $d$. Write $n$ in the form $dk+m$ – user51427 Dec 7 '12 at 22:02

Note that $n(n+1)\dots(n+d-1)$ is the product of $d$ consecutive integers. Thus, it suffices to prove that if $n,n+1,\dots,n+d-1$ are any $d$ consecutive integers, then $d$ divides one of these integers. I would prove this by induction on $n$, simultaneously for all $d$.

First, it’s clearly true for $n=1$, since in that case the $d$ integers are $1,2,\dots,d$. Suppose now that it’s true for some $n$, and consider the $d$ consecutive integers $n+1,n+2,\dots,n+d$. By the induction hypothesis one of the integers $n,n+1,\dots,n+d-1$ is a multiple of $d$. If it’s one of the integers $n+1,n+2\dots,n+d-1$, you’re done (why?), and if $n$ is a multiple of $d$, then so is ... ?

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Hint $\$ Any sequence of $\rm\,d\,$ consecutive naturals has an element divisible by $\rm\,d.\,$ This has a very simple proof by induction: shifting such a sequence by one does not change its set of remainders mod $\rm\,d,\,$ since it simply replaces the old least element $\rm\:\color{#C00}n\:$ by the new greatest element $\rm\,\color{#C00}{n+d}$

$$\begin{array}{}\rm \:\color{#C00}n &\rm n+1 &\rm n+2 &\rm\! \cdots\! &\rm n+d-1 & \\ \to &\rm n+1 &\rm n+2\rm &\! \cdots\! &\rm n+d-1 &\rm \color{#C00}{n+d} \end{array}$$

Since $\rm\: \color{#C00}n\equiv \color{#C00}{n+d} \pmod{\! d},\,$ the shift does not change the remainders in the sequence. Thus the remainders are the same as the base case $\rm\ 0,1,2,\ldots,d-1\ =\:$ all possible remainders mod $\rm\,d.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $\rm\,d.$

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Hint

$$(n+1)(n+2)...(n+d-1)(n+d)= \left[(n+1)(n+2)(n+3)...(n+d-1)\right]n + \left[(n+1)(n+2)(n+3)...(n+d-1) \right]d$$

$P(n)$ tells you that the first term on RHS is divisible by $d$, while the second one is clearly divisible by $d$...

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Typo: there are two factors $(n+1)$ in each term. – WimC Dec 7 '12 at 22:09
@WimC Thank you, fixed. – N. S. Dec 7 '12 at 22:27

Fixed $d$, it is easy to show that the statement holds for every $n$, by induction. The inductive step is as follows: suppose $d|n(n+1)\cdots(n+d-1)$, then put $m=gcd(d,n)$ and observe that $gcd(d,n+d)=m$ as well, therefore, by assumption $$\frac{d}{m}\vert (n+1)\cdots(n+d-1)$$ so $$d=\frac{d}{m}m\vert (n+1)\cdots(n+d-1)(n+d)\;.$$ Now you just have to show that the statement is true for $n=2$ and every $d$, but this is easy: $2\leq d\leq d+2-1=d+1$, therefore $d$ divides $2\cdot3\cdots(d+1)=(d+1)!$.

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This is one that iterates on the relative values between n & d for the inductive steps-- not on values n, n+1, n+k, n+k+1 as the usual case.

The proof is by looking at the values of n and d. Say n=dt+x whenever n>d.

For the base case, check whether x=0 satisfies-- which it does.

For the inductive step, assume k=x satisfies. Then, so does k=x+1 since the multipliers are the same.

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As abbreviation, define $f(n,d)=n\cdot (n+1)\cdot\ldots\cdot (n+d-1)$.

Base case $n=2$: $$\forall d\ge 2\colon d|f(2,d) = 2\cdot 3\cdot\ldots \cdot d$$ This is true because $d$ occurs among the factors on the rihght.

$n\to n+1$:

Assume that we know about $n$ that $$\tag1\forall d\ge 2\colon d|f(n,d)$$ Claim: then also $$\tag2\forall d\ge 2\colon d|f(n+1,d)$$ Indeed, let $d\ge 2$ be arbitrary. Then by$(1)$ we know that $d|f(n,d)$, say $f(n,d)=d a$. Note that $$n\cdot f(n+1,d) = f(n,d)\cdot (n+d)=n\cdot f(n,d)+d\cdot f(n,d)$$ Since $n$ divides $f(n,d)$ (as first factor), write $f(n,d)=n b$. Then we obtain $$f(n+1,d) = \frac1n(n\cdot f(n,d)+d\cdot f(n,d))=d (a+b).$$ SInce $d\ge 2$ was arbitrary, this shows $(2)$ and thus our induction step $n\to n+1$.

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Consider the numbers $n \mod d,(n+1)\mod d,...,(n+d-1)\mod d$ these numbers are all equal and form a subset of {$0,1,...,d-1$}. Since both sets are equal on size, therefore both sets are equal. Therefore one of the numbers $n,n+1,...,n+d$ is divisible by $d$. Thus, $d|n(n+1)...(n+d-1)$

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