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Let's $n \in \mathbb{Z^+}$, how to $\text{prove}|\text{disprove}$ that:

the equation $\boxed{\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0}$ has exactly $n$ distinct negative roots.

My friend get bored, then he started play with equations. And he found that $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ with $n = 1, 2, 3$ have $1, 2, 3$ distinct negative roots respectively, he asked me to prove the "quoted question" above; but seem like I can't do it, please help or give me a hint.

Thank you. Please help me edit my post & tags, I am not good at English

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Xeing
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    I'd like to have a friend like that, nice question! :) I think the expression for the polynomial can be simplified to $\sum_{i=0}^n \binom{n}{i}^2x^{i}$, since the binomial coefficients are symmetric. – Daniel R Dec 18 '13 at 13:16
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    Minor remark: the fact that the roots must be negative is obvious since the coefficients are all positive. A second minor remark: almost all polynomials have distinct roots and so does probably yours as well. It might be very hard to prove it but unless something special is going on (like the binomial formula for $(1+x)^n$) it's really not surprising, or even interesting. You could ask similar question for a polynomial having coefficients given by any function of binomial numbers whatsoever (except the trivial one I just mentioned). – Marek Dec 18 '13 at 13:49
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    @Marek It may be that "almost all polynomials have distinct roots" (i.e. are separable), but it is NOT true that almost all polynomials have distinct real roots. That's what makes this question interesting. – Caleb Stanford Dec 18 '13 at 14:00
  • @Marek Consider, for example, monic quadratic polynomials $x^2 + bx + c$. These have real roots for all pairs $(b,c) \in \mathbb{R}^2$ with $c \le \frac14 b^2$. This is not "almost all" polynomials. – Caleb Stanford Dec 18 '13 at 14:02
  • @Goos: fair enough. But I don't think the interesting part of the question is in proving the reality of the roots. – Marek Dec 18 '13 at 14:36
  • @Marek A polynomial with positive coeffs need not have negative roots. $x^2 + 1 = 0$ has no negative root. – DanielV Dec 18 '13 at 15:10
  • @DanielV: thank you Captain Obvious. Since the question talks about negativity of roots it obviously assumes they are real (since there is no notion of negativity on complex roots, there being no order on complex numbers). If this is not so then the question's title is misleading and should state that OP wants to know whether the roots are real. – Marek Dec 18 '13 at 15:22
  • @Marek what is the interesting part of the question if it isn't proving the roots are real? As you state yourself, it's clear that all the real roots of the polynomial must be negative. It's also isn't too weird to expect all the (possibly complex) roots to be distinct, which you also pointed out. Therefore the only thing left to be interesting is proving all the roots are real. – Caleb Stanford Dec 18 '13 at 15:33
  • @DanielV Note that Marek was saying NOT that any polynomial with positive coefficients has a negative root, but that all real roots of such a polynomial will be negative. However, Marek's response to you is confusing me a bit. – Caleb Stanford Dec 18 '13 at 15:35
  • @Marek I understand now I think how you were confused. But it is normal to refer to a complex root which is a negative real number as a "negative root". Likewise, if I told you to prove a certain real number was a multiple of 2, you would also have to prove it was an integer. – Caleb Stanford Dec 18 '13 at 15:38
  • @Goos: there is nothing really interesting in this question as I said. One can asks dozen similar questions. But the accidental connection to Legendre polynomial pointed out in Ragib's answer is neat, so that's the only interesting thing going on here. – Marek Dec 18 '13 at 15:42
  • @Goos: if it's a standard usage then be it (I don't think it is) but I still need not like it. Same thing with that multiple of $2$. It's really two distinct questions, one for integrality and one for divisibility by $2$. Conflating them into one question doesn't really do anyone a service. You save about three words at the expense of possibly confusing anyone who doesn't follow exact same conventions you do. – Marek Dec 18 '13 at 15:44
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    @Marek, it's confusing at first but it definitely is standard usage. For example if I write $\lim_{x \to 1} x^2 = 1$, I am actually saying two things: first that the limit exists, and second that the limit equals $1$. In the same way, it is implied when I say "prove the roots are negative" that you also need to prove they are real. Specifying this all the time would be a pain. – Caleb Stanford Dec 18 '13 at 15:46
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    @Goos: we can agree to disagree. It largely depends on context and audience when you should conflate two problems into one (for example when teaching limits for the first time, you should definitely not conflate the two). I for one prefer to be explicit and don't consider making myself clear a pain. Rather I find it painful if someone doesn't take the effort in presentation and assume everybody thinks the same way they do and knows the same facts as well. If you want to continue the discussion let's move it into chat, I think we spammed this thread enough :) – Marek Dec 18 '13 at 15:52

1 Answers1

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The polynomial in question is equal to $(1-x)^n P_n \left( \dfrac{1+x}{1-x} \right)$ where $P_n$ is the $n$-th Legendre polynomial. So the question becomes boils down to the fact that $P_n$ has $n$ distinct roots in $(-1,1),$ which is proved here.

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