Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Recently, I wondered about the following problem: let $n\geq 5$ and let

$$ P_n(x)=(x-1)(x-2)\ldots (x-n)-1 $$

Is it true that $P_n(x)$ has $n$ distinct real roots for any $n\geq 5$ ? I checked it for up to $n=50$. I found nothing else so far.

share|cite|improve this question
The same polynomial appears in this question. – Jyrki Lahtonen Jun 12 '14 at 19:19
When thinking for a solution to this, a problem came up. How would I differentiate this for any n? – Cruncher Jun 12 '14 at 19:36
The problem is equivalent to find $x$ in $\Gamma(x)=\Gamma(x-n)$ because $P_n(x)=-1+\dfrac{\Gamma(x)}{\Gamma(x-n)}$. – Jika Jun 12 '14 at 19:58

What happens when you evaluate this at $x=0, 1, 1.5, 2, 2.5 \dots$? Then use the intermediate value theorem.

At the integer points the polynomial evaluates to $-1$. At alternate intermediate points the value is positive, and you get a pair of roots.

If $n$ is even you have positive values at $x=0$ and $x=n+1$ so roots at "either end" to make the full complement.

If $n$ is odd you have a positive value at $x=n+1$ which does the same.

share|cite|improve this answer

Let $Q_n(x) = P_n(x) +1 $. Chect the values at $Q_n(x)$ at points $\frac{i}{2}$ for $i = 1,3,5...,2n-1$. The signs will alternate and the value of $|Q_n(x)|$ will be greater than $1$. So, the value of $P_n(x)$ will also alternate at these points. This means that there are $n$ roots, in between these points.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.