<s>[INST] Product of consecutive natural numbers Let $n,k$ be two natural numbers, $n\ge k+2$ and 
$$(n-k-1)(n-k)(n-k+1)(n-k+2)=k(k+1)n(n+1).$$
In the LHS we have product of 4  consecutive  numbers  and in the RHS  we have the  product of 2 consecutive  numbers  on the  product of 2 consecutive  numbers.
Problem. Prove that  $k,k+1,n,n+1$  also are   consecutive  numbers.
My  attempt.  All factors are in increasing order. Suppose  that $n-k-1=k$, then $n=2k+1$ and on we can reduce the products to $k(k+1)(k+2)(k+3)=k(k+1)n(n+1)$ or $(k+2)(k+3)=n(n+1)=(2k+1)(2k+2)$   We have  equal  products of two 2 consecutive  numbers  so it implies that $k+2=2k+1$ and  we get a solution  $k=1, n=3$. 
How  to prove that the case $n-k-1 \neq k$ is impossible?
 [/INST] Let $a:=n-k\ (\ge 2)$.
Then, we have
$$(a-1)a(a+1)(a+2)=k(k+1)(k+a)(k+a+1)$$
This can be written as
$$(2k^2+2ka+2k+a)^2=4a^4+8a^3-3a^2-8a$$
Now, for $a\gt 2$, we have
$$(2a^2+2a-2)^2\lt 4a^4+8a^3-3a^2-8a\lt (2a^2+2a-1)^2$$
from which we have that $4a^4+8a^3-3a^2-8a$ cannot be a square number for $a\gt 2$.
So, we have to have $n-k=a=2$.
It follows from this that $k,k+1,n,n+1$ are consecutive numbers.
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