We say that the number $N \in \mathbb{N}$ has the property $P(k)$ if it can be factored into a product of $k$ consecutive natural numbers (not equal to $1$).
Find the value of $k$ such that some $N$ simultaneously has the properties $P(k)$ and $P(k+2)$.
This is a problem from a Soviet high-school math contest (1981).
The book I found it in only offers the following answer:
$$k=3;~~720=2 \cdot 3 \cdot 4 \cdot 5 \cdot 6=8 \cdot 9 \cdot 10$$
Is this the only solution?
How can this problem be solved in general? (Using high-school level math)
My attempt. For some $n(N,k), m(N,k) \in \mathbb{N}$ we have:
$$N=\frac{n!}{(n-k)!}$$
$$N=\frac{m!}{(m-k-2)!}$$
Obviously:
$$m<n$$
So it seems to me we have a three variable equation which needs to be solved in natural numbers $m,n,k$:
$$\frac{n!}{m!}=\frac{(n-k)!}{(m-k-2)!}$$
I'm not sure how this can be solved.
I suppose we can also write:
$$\frac{N}{k!}=\left( \begin{array}( n \\ k \end{array} \right)$$
$$\frac{N}{(k+2)!}=\left( \begin{array}( ~~~m \\ k+2 \end{array} \right)$$
So:
$$(k+1)(k+2)\left( \begin{array}( ~~~m \\ k+2 \end{array} \right)=\left( \begin{array}( n \\ k \end{array} \right)$$
Edit
By abusing Mathematica I found several other solutions (see also @stewbasic's comment):
$$k=20,~~~~n=24,~~~~m=23,~~~~N=25852016738884976640000$$
$$k=55,~~~~n=60,~~~~m=59,~~~~N= \text{a very big number}$$
I wonder, do all the solutions except for $k=3$ and $k=4$ obey the rule:
$$n=m+1$$
For this case we can easily write:
$$m+1=\frac{(m-k+1)!}{(m-k-2)!}=(m-k-1)(m-k)(m-k+1)$$
So we get a Diophantine equation:
$$ m^3- 3 k m^2+ 3 k^2 m - 2 m-1 + k - k^3=0$$
$$(m-k)^3=2m+1-k$$
This one has many solutions, it seems, for example:
$$k=328,~~~~m=335$$
$$k=495,~~~~m=503$$
$$k=710,~~~~m=719$$
Al of this doesn't answer my question - how high-schoolers were supposed to solve this problem?
Also, is there only one $N$ for each $k$ in general? How to show this if true?
Integer solutions of your equation (in the case $n=m+1$) $(m-k)^3 = 2m+1-k$ are obtained by solving a system of linear equations in $m$ and $k$: $$ \eqalign{t &= m-k\cr t^3 &= 2m + 1 - k\cr}$$ to get $$ m = t^3 - t - 1, \ k = t^3 - 2 t - 1 $$ This gives you positive integer values of $m$ and $k$ iff $t$ is an integer $\ge 2$.