How to prove that the product of eight consecutive numbers can't be a number raised to exponent 4?

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.

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.