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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!

Math_Day
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1 Answers1

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(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.
Calvin Lin
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