# How many ways are there to prove that the product of 8 consecutive positive integers is never a square?

I saw this on quora. Here is my solution. (I know about general theorems (Erdos, Erdos-Strauss) that imply this. I wanted a proof that worked for only this special case (8 integers, square).)

All computations done by Wolfy.

$$(n^4+14n^3+63n^2+98n+27)^2 \lt \prod_{k=0}^7 (n+k) \lt (n^4+14n^3+63n^2+98n+28)^2$$ for $$n \ge 4$$. Therefore $$\prod_{k=0}^7 (n+k)$$ can not be a square for $$n \ge 4$$.

For $$1 \le n \le 3, \prod_{k=0}^7 (n+k)$$ contains exactly one 7 and so can not be a square.

That magic polynomial was gotten from $$\sqrt{\prod_{k=1}^7 (1+kx)} = 1 + 14 x + 63 x^2 + 98 x^3 + 28 x^4 + O(x^5)$$.

What other ways are there to prove this?

Are there any proofs that just use divisibility? (I looked at the power of 2 that divided the product, but that can be even.)