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. 
 A: 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.
A: Here's a proof that there is no positive integer $n$ such that
$$f(n) = (n)\cdots (n+7)$$
is a perfect $4$-th power. 

The proof is along the same lines as the proof I gave showing that there is no positive integrer $n$ such that $f(n)$ is a perfect square, but the proof for the case of $4$-th powers is easier, and can be done by hand.

Thus, suppose $n$ is a positive integer such that $f(n)$ is a perfect $4$-th power.

We start by getting a quick lower bound on $n$ . . .

Considering factors of $7$, it's clear that $f(n)$ must be a multiple of $7$, hence must be a multiple of $7^4$.

But at most two of the $8$ factors can be multiples of $7$, and at most one of them can be a multiple of $7^2$. Thus, in order for $f(n)$ to be a multiple of $7^4$, one of the $8$ factors must be a multiple of $7^3$.

It follows that $7^3 \le n + 7$, so $n \ge 336$.

Next we pair up the $8$ factors . . .

Since
\begin{align*}
n(n+7) &= n^2 + 7n\\[4pt]
(n+1)(n+6) &= n^2 + 7n + 6\\[4pt]
(n+2)(n+5) &= n^2 + 7n + 10\\[4pt]
(n+3)(n+4) &= n^2 + 7n + 12\\[4pt]
\end{align*}
it follows that
$$(n^2 + 7n)^4 < f(n) < (n^2 + 7n + 12)^4$$
hence we must have
$$f(n) = (n^2 + 7n + c)^4$$
for some positive integer $c < 12$.

From the definition of $f(n)$, we have $\bigl((n)(n+7)\bigr) \mid f(n)$.
\begin{align*}
\text{Then}\;\;&\bigl((n)(n+7)\bigr) \mid f(n)\\[4pt]
\implies\; &f(n) \equiv 0 \pmod {n^2 + 7n}\\[4pt]
\implies\; &(n^2 + 7n + c)^4 \equiv 0 \pmod {n^2 + 7n}\\[4pt]
\implies\; &c^4 \equiv 0 \pmod {n^2 + 7n}\\[4pt]
\implies\; &n^2 + 7n \le c^4\\[4pt]
\implies\; &n^2 < c^4\\[4pt]
\implies\; &n < c^2\\[4pt]
\implies\; &n < 144\qquad\text{[since $c < 12$]}\\[4pt]
\end{align*}
contrary to our lower bound, $n \ge 336$.

Hence there is no positive integer $n$ such that $f(n)$ is a perfect $4$-th power.
