Is there any elementary proof for the statement”the product of any eight consecutive positive integers is never a perfect square.”? In 2004, my students discovered that the consecutive products are never a square for the length of n=2,3 and 4, they were very interested in it and started their investigation on consecutive products of lengths 5 or above. After investigating the consecutive products for months, some of them found a paper in 1975 written by P. ERDOS and J.L. SELFRIDGE stated that
$\text{THE PRODUCT OF CONSECUTIVE INTEGERS IS NEVER A POWER} \tag*{}$
After knowing this ‘bad’ news, they didn’t stop and continued their investigation. They have found many fruitful results for which I am very proud of them. Some results are elementary but so beautiful that I want to share with you all.
One of my students studying secondary 3 invented a method on his own to prove that any consecutive products of length 8 is lying between 2 consecutive squares and hence is never a square. He first split the consecutive product into 2 products of polynomials $M$ and $N$ of ‘near size‘ and then obtained such beautiful inequalities
$\left(\frac{M+N}{2}-1\right)^{2}< M \times N<\left(\frac{M+N}{2}\right)^{2}. \tag*{}$
and made his final conclusion!
My question
Is there other elementary method to prove that product of eight(or even $n$) positive integers is never a perfect square?
 A: Proof:
Step 1(Splitting the consecutive product into M and N)
Let the consecutive positive integers be $$(n-3),(n-2),(n-1), n,(n+1),(n+2),(n+3) \text { and }(n+4), $$
$$M=(n-3) n(n+2)(n+3)=n^{4}+2 n^{3}-9 n^{2}-18 n, $$
and
$$N=(n-2)(n-1)(n+1)(n+4)=n^{4}+2 n^{3}-9 n^{2}-2 n+8 $$
Then
$$M \times N=n^{8}+4 n^{7}-14 n^{6}-56 n^{5}+49 n^{4}+196 n^{3}-36 n^{2}-144 n ,$$
$M$ and $N$ so formed are of ’near size‘ in the sense that they have the same coefficients of higher terms.
Step 2(Investigation on the square of (M+ N)/2)
$\left(\frac{M+N}{2}\right)^{2}$
$=\left(n^{4}+2 n^{3}-9 n^{2}-10 n+4\right)^{2}$
$=n^{8}+4n^{7}-14n^{6}-56n^{5}+49n^{4}+196n^{3}+28n^{2}-80n+16 $
$>M \times N\cdots(1)$
Step 3(Investigation on the square of (M+N)/2-1)
On the other hand,
$$\because \left(\frac{M+N}{2}-1\right)^{2}=n^{8}+4 n^{7}-14 n^{6}-56 n^{5}+47 n^{4}+192 n^{3}+46 n^{2}-60 n+9.\\$$
$$\therefore M \times N-\left(\frac{M+N}{2}-1\right)^{2} = 2 n^{4}+4 n^{3}-82 n^{2}-84n-9$$
He got stuck here and thought : How to prove that $2 n^{4}+4 n^{3}-82 n^{2}-84n-9>0$?
Later he was advised by me to try to divide this polynomial by $n-6$ and get
$\quad 2 n^{4}+4 n^{3}-82 n^{2}-84n-9$
$=(n-6)\left(2 n^{3}+16 n^{2}+14 n\right)-9$
$>0 \quad \text { for } n \geq 7,$
We obtained another inequality (2):
$M \times N>\left(\frac{M+N}{2}-1\right)^{2} \tag*{(2)}$
Conclusion(no square lying between 2 consecutive squares)
Combining the two inequalities $(1)$ and $(2)$, when $n \geq 7$, we get
$\left(\frac{M+N}{2}-1\right)^{2}< M \times N<\left(\frac{M+N}{2}\right)^{2}.  \tag*{}$
Since $M \times N$ lies between two consecutive squares, $M \times N$ must not be a square. Moreover,
$1 \cdot 2 \cdot 3 \cdots \cdot 8,\quad 2 \cdot 3 \cdot 4 \cdot \cdots 9 $
$\text { and } 3 \cdot 4 \cdot 5 \cdots 10 $are not squares. Therefore the product of eight consecutive positive integers is never a square.
