Square of product of successive naturals

I’ve been working on an interesting task from a regional math contest:

Prove that the product of 8 successive naturals cannot be a natural number to the power of 4.

To prove this, we will first take a look at two other theorems (and prove them), and then use them to prove the original statement.

I. \sqrt{x} irrational \implies \sqrt{\sqrt{x}} irrational

To prove this, it suffices proving the contrapositive:
\sqrt{\sqrt{x}} rational \implies \sqrt{x} rational.

We have that for some a, b, a/b = \sqrt{\sqrt{x}}

Square both sides to get a^2/b^2 = \sqrt{x}. Thus, \sqrt{x} is rational.

II. \sqrt{x(x+1)} is irrational

To prove this, note that x and (x+1) need to be squares. Consider for some a, x = a^2. Further, for some b, x + 1 = b^2.

Now, b^2 - a^2 = (b - a)(b + a) = 1. But the only way this is possible if b - a = 1/(b + a).

Since a and b are positive naturals, we reach a contradiction for the identity above and thus either x or x+1 have no squares. In either case, \sqrt{x(x+1)} is irrational.

III. Prove that there is no y s.t. y^4 = x(x+1)(x+2)(x+3)(x+4)(x+5)(x+6)(x+7) = *

We will assume that such y exists and reach a contradiction.

Rewrite as y = \sqrt{\sqrt{*}} and suppose y is rational.

From I we have that it suffices to only prove that \sqrt{*} is rational.

From II we have that either x or (x+1) is irrational. At least one of the 8 elements has no square, and we reach a contradiction. Thus y is irrational.

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