# 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.