<s>[INST] How to prove that the product of eight consecutive numbers can't be a number raised to exponent 4? How to prove this? I tried something like $$P(n,8)=\frac{n!}{(n-8)!} = b^4$$ but I can't proceed to a solution. 
 [/INST] Let 
$$f(x) = (x)\cdots(x+7)$$
$$g(x) = x^2 + 3x + 1$$
$$h(x) = x^2 + 11x + 29$$
Then for $x > 0$,
$$f(x) < (g(x)h(x) - 1)^2$$
and for $x \ge 4$,
$$(g(x)h(x)-2)^2  < f(x)$$
hence, if $n$ is a positive integer with $n \ge 4$,
$$(g(n)h(n)-2)^2 < f(n) < (g(n)h(n)-1)^2$$
so $f(n)$ can't be a perfect square.

It's easily verified that $f(1),f(2),f(3)$ are not perfect squares (since for example, they're multiples of $7$, but not multiples of $7^2$).

Therefore the product of $8$ consecutive positive integers can't be a perfect square.

Some explanation of where $g,h$ came from . . .

Identically,
\begin{align*}
x(x+1)(x+2)(x+3) &= \bigl((x)(x+3)\bigr)\bigl((x+1)(x+2)\bigr)\\[4pt]
&=(x^2+3x)(x^2+3x+2)\\[4pt]
&=\bigl((x^2+3x+1)-1)\bigr)\bigl((x^2+3x+1)+1)\bigr)\\[4pt]
&=(x^2+3x+1)^2-1\\[4pt]
\end{align*}
Then, replacing $x$ by $x+4$,
$$(x+4)(x+5)(x+6)(x+7) = (x^2+11x+29)^2-1
\qquad\qquad\qquad\qquad\qquad\;\;\;$$
Thus, $f(x)$ is algebraically "close" to
$$(g(x)h(x))^2$$
which motivates the idea of trying to "trap" $f(x)$ between two nearby perfect squares.
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