$n_1, n_2,...,n_s$ such $(n_1+k)(n_2+k)\cdots(n_s+k)$ integral multiple of $n_1n_2\cdots n_s$. Prove $|n_i| = 1$ for some $i$ Let $n_1, n_2,...,n_s$ be distinct integers such that
\begin{equation}
    (n_1+k)(n_2+k)\cdots(n_s+k)
\end{equation}
is an integral multiple of $n_1n_2\cdots n_s$ for every integer $k$. Give a proof or a counterexample of the following statement: $|n_i| = 1$ for some $i$
I think this statement is always true. The way decided to approach was to arrange $n_1,...,n_s$ such that $|n_1|\geq |n_2|\geq \cdots \geq|n_s|$. I was hoping to plug in $k = \pm 1$ based on the sign of $n_2$ but I didn't get very far past that.
Some hints would be greatly appreciated!
 A: (Fill in the gaps as needed, esp when it says "Show that". If you're stuck, show your work and explain what you've tried.)
Hint: Show that $n_i \neq 0$, so henceforth $n_i \neq 0$.
Let $f(k) = \prod ( n_i + k)$.
Proof by contradiction. Suppose $ |n_i | \neq 1$.
Show that $f(0), f(1), f(-1) \neq 0$.
Hint: Show that $ 0 < f(1) f(-1) < f(0)^2$.
Hence, it is not possible for both $ f(0) \mid f(1) $ and $f(0) \mid f(-1)$, contradicting the condition that $f(0) \mid f(k)$.

Notes for how I came up with this solution:

*

*It's initially not clear how to use the condition that $ f(0) \mid f(k)$.

*The condition that $|n_i| \neq 1$ suggests that we look at $ f(1)$, $f(-1) \neq 0 $.

*This suggests that we have either $ f(0) \not \mid f(1)$ or $f(0) \not \mid f(-1)$.

*We then have to figure out how this can be proven. We would normally have very little control over these terms. For example, it is not generally true that $ f(1) f(-1) < f(0) ^2$.

*This happens only if $ n_ i \neq 0$, so let's investigate what happens if some $n_i = 0$.

*Thankfully, $n_i = 0$ isn't allowed by the condition, so now we have $f(1) f(-1) < f(0)^2$.

*If so, one simple way to show that $ f(0) \not \mid f(1), f(-1)$ is that $ 0 < |f(1) f(-1)| < f(0)^2$, and we can verify that is true.

