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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.)

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