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Let $P(x) \in \mathbb{R}[x]$ be polynomial with all real roots and has the property that $P(a)=0 \Rightarrow P(a+1)=1$ for all $a \in \mathbb{R}$. Prove that $P(x)$ has a repeated root.

I think this problem statement is not true because if we suppose that $P(x)=x$ then $P(0)=0 \Rightarrow P(1)=1$, $\;P(x)$ has no repeated root.

Please suggest.

Paolo Leonetti
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user403160
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3 Answers3

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Let us suppose that $P \in \mathbb{C}[x]$ has degree $k$, i.e., it has complex roots $\alpha_1,\ldots,\alpha_k$ with $$ P(x)=c\prod_i(x-\alpha_i) $$ where the $\alpha_i$ are distinct complex and $c\neq 0$. In addition, we know that $$ P(x)-1=c\prod_i(x-\alpha_i-1). $$ In particular, the coefficient of $x^{k-1}$ of $P(x)$ verifies $$ -c\sum_i \alpha_i = - c\sum_i (\alpha_i+1). $$ This is impossible whenever $k\ge 2$.

Edit: The condition of $\alpha_i$ being distinct is necessary. Indeed, the polynomial $P(x):=x^k$ has all roots real and equal to $0$, and $P(a)=0 \implies P(a+1)=1$ for all $a \in \mathbf{R}$.

Paolo Leonetti
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  • The last equality is just a particular instance $[x^i]P(x)=x^i$ which holds whenever $i>\mathrm{deg}(Q(x))$. – Paolo Leonetti Aug 07 '17 at 11:50
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    Why is there no contradiction if a repeated root exists? – user236182 Aug 07 '17 at 12:18
  • There is still a contradiction if a repeated root exists. But a false statement implies anything. So we can still claim a repeated root exists. https://www.google.com/search?site=&source=hp&ei=31uIWZ3aHpHQwAK0n62ICQ&q=false+statement+implies+anything&oq=false+statement+implies&gs_l=mobile-gws-hp.1.0.0i22i30k1.7905.55358.0.55933.31.26.3.1.1.0.193.3860.0j25.25.0....0...1.1j4.64.mobile-gws-hp..2.29.3934.3..0j5j35i39k1j0i131k1j0i19k1.yYdvHlDwYwk – user236182 Aug 07 '17 at 12:31
  • @user236182 $x^2$ is the counterexample – Paolo Leonetti Aug 07 '17 at 13:43
  • @PaoloLeonetti you have misunderstood a little. First of all the roots are real. The problem doesn't say that if a polynomial has multiple roots $\zeta$ then $P(\zeta+1)=1$ but the reverse. If it has roots such that $P(\zeta+1)=1$ then $\zeta$ is a multiple root – Raffaele Aug 07 '17 at 14:56
  • @Raffaele I just considered the more general question where $P(x)$ has complex coefficient and complex distinct roots; and, for each complex root $\alpha$, then $\alpha+1$ is a root of $P(x)-1$. Btw, your answer above appears to me quite nonsense. – Paolo Leonetti Aug 07 '17 at 15:01
  • @PaoloLeonetti Nonsense because it works? Take $P(x)=(x-1)^3$ and compute $P(2)$ – Raffaele Aug 07 '17 at 15:28
  • @Raffaele ..indeed $1$ is a multiple root of $P$ and you confirming the result just for that choice of $P$. – Paolo Leonetti Aug 07 '17 at 15:34
  • @PaoloLeonetti No no no. I have proved what I am saying. I'll add to my answer – Raffaele Aug 07 '17 at 15:41
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    I get the idea from Paolo Leonetti's general solution and I do my work for real number. Thank you very much for all your help, Paolo Leonetti, Lord Shark the Unknown and Raffalel.:) – user403160 Aug 07 '17 at 16:57
  • @Raffaele Please check the comment at your answer – Paolo Leonetti Aug 08 '17 at 23:44
  • @carat You're welcome – Paolo Leonetti Aug 08 '17 at 23:44
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You need the degree of $P$ to be $d\ge2$. If so then $$P(x)=(x-a_1)(x-a_2)\ldots(x-a_d)$$ for distinct $a_i$. Then $$P(x)-1=(x-a_1-1)(x-a_2-1)\ldots(x-a_d-1).$$ Can you get a contradiction from these?

Angina Seng
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For $n>1$ we have $P(x)=(x-a)^n$ then $P(a+1)=(a+1-a)^n=1$

If $P(x)$ has no repeated root then it doesn't happen in general

Anyway carat, the original poster, is right: for $n=1$ it is false. It even makes no sense talking of multiple root for $n=1$

Suppose the root are $3$ then we have

$P(x)=(x - a) (x - b) (x - c)$

Use the condition that $P(a+1)=1;\;P(b+1)=1;\;P(c+1)=1$

$$ \left\{ \begin{gathered} \left( {a + 1 - a} \right)\left( {a + 1 - b} \right)\left( {a + 1 - c} \right) = 1 \hfill \\ \left( {b + 1 - a} \right)\left( {b + 1 - b} \right)\left( {b + 1 - c} \right) = 1 \hfill \\ \left( {c + 1 - a} \right)\left( {c + 1 - b} \right)\left( {c + 1 - c} \right) = 1 \hfill \\ \end{gathered} \right. $$ which simplifies to $$\left\{ \begin{gathered} \left( {a - b + 1} \right)\left( {a - c + 1} \right) = 1 \hfill \\ \left( {b - a + 1} \right)\left( {b - c + 1} \right) = 1 \hfill \\ \left( {c - a + 1} \right)\left( {c - b + 1} \right) = 1 \hfill \\ \end{gathered} \right.$$

which is verified when $a=b=c$

Therefore $x=a$ is a triple root. $P(x)=(x-a)^3$

In a similar way it can be proved for any degree

Raffaele
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  • Ok you got a system of three non linear equations with three unknowns. In principle you do not know whether there are one, two, or infinitely many solutions. You have to prove that, for each possible solution, at least two unknowns are equal.. Now? – Paolo Leonetti Aug 07 '17 at 19:19