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It seems that if you square the coefficients in the expansion of $(x+1)^n$, the resulting polynomial has $n$ distinct real roots. (I experimented using desmos.com.)

In other words, I am looking for a proof (or counter-example) for the proposition that polynomial $\displaystyle p(x) = \sum_{k=0}^n\binom{n}{k}^2x^k$ has exactly $n$ distinct real roots.

But I have not found a proof. I tried induction, to no avail.

(Context: I am looking for an expanded polynomial with exactly, say, $100$ distinct real roots.)

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Jul 10, 2021 at 13:44

4 Answers 4

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We have $p(x) = (1-x)^n P_n\left(\frac{1+x}{1-x}\right)$ where $P_n$ is the Legendre polynomial of degree $n$. This follows from the Rodrigues formula, among other ways. (Mathematica spotted it for me.) The Legendre polynomials are orthogonal polynomials, so by standard theory their roots interlace each other, and in particular $P_n(x)$ has $n$ distinct real roots. Moreover it is easy to see $-1$ is not a root.

The function $x \mapsto \frac{1+x}{1-x}$ is bijective from $\mathbb{R} \setminus \{1\}$ to $\mathbb{R} \setminus \{-1\}$. Hence the $n$ distinct roots of $P_n$ are hit by $n$ distinct values in $\mathbb{R} \setminus \{1\}$, so $p(x)$ has at least, hence exactly, $n$ distinct real roots.

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  • $\begingroup$ Wow, that was amazing. $\endgroup$
    – Trebor
    Jul 9, 2021 at 9:37
  • $\begingroup$ (Now I see @TeresaLisbon gave the same argument in the rather extended comments discussion on the main post.) $\endgroup$ Jul 9, 2021 at 9:39
  • $\begingroup$ @JoshuaP.Swanson Thanks , yes , I knew it would be the case. We can try a standalone argument, but it seems that is not required now. $\endgroup$ Jul 9, 2021 at 9:40
  • $\begingroup$ @TeresaLisbon I briefly considered doing just that--making a completely self-contained argument--but moments later it proved too tedious for me to bother. If anybody wants to do so, they're welcome to edit my answer or post their own. $\endgroup$ Jul 9, 2021 at 9:41
  • $\begingroup$ @JoshuaP.Swanson True, it was too tedious! $\endgroup$ Jul 9, 2021 at 9:46
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NOT A FULL ANSWER $$p_n(x)=\sum^n_{k=0} {n\choose k}^2x^k$$ for a polynomial to have a triple root, we need $p'$ and $p$ to share a factor: $$p_n'(x) = \sum^n_{k=0} {n\choose k}^2\cdot k\cdot x^{k-1}$$ $$p_n'(x) = \sum^n_{k=0} {n\choose k}^2\cdot \frac{k}{n}\cdot n\cdot x^{k-1}$$ Note that ${n \choose k} = \frac{n}{k} {n-1 \choose k-1}$ $$p_n'(x) = n\sum^n_{k=0} {n-1\choose k-1}{n\choose k}x^{k-1} =n\sum^n_{k=0} {n-1\choose k-1}({n-1 \choose k}+{n-1 \choose k-1})x^{k-1}=np_{n-1}(x)+ n\sum^n_{k=0} {n-1\choose k-1}{n-1\choose k}x^{k-1}$$ $$\overbrace{np_n(x)}^{\text{multiple of } p_n}-nx^n+n\sum^n_{k=0} {n-1\choose k-1}{n-1\choose k}x^{k-1}$$

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  • $\begingroup$ You lost a factor of $n/k$. $\endgroup$
    – Trebor
    Jul 9, 2021 at 7:45
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This is not an answer.

$$p_n(x) = \sum_{k=0}^n \binom{n}{k}^2 x^k = \sum_{k=0}^n \binom{n-1}{k-1}^2 x^{k-1}\cdot \frac{n^2x}{k^2}$$

So to relate this to $p_{n-1}$, we need to differentiate, multiply by $x$, and differentiate:

$$(xp_n'(x))' = \sum_{k=0}^{n-1}\binom {n-1}k^2 x^k n^2 = n^2 p_{n-1}(x)$$

So $p'_n(x) + xp''_n(x) = n^2 p_{n-1}(x)$. We can pack this up into a generating function:

$$P(x,y) = \sum_{n}p_n(x) y^n.$$

And $P_x + x P_{xx} = P_y + y P_{yy}$, which is interesting.

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This is not very well polished, and seems to contain some errors. But I believe it is workable into a complete proof.

We want to let one term dominate all the others.

$\binom nk^2 x^k \over \binom {n}{k-1}^2 x^{k-1}$ is equal to $(\frac{n-k+1}{k})^2x$. Dub $a_k = \binom nk^2 x^k$. Then $a_{k-1} : a_k : a_{k+1} = \frac{k^2}{(n-k+1)^2x}:1:\frac{(n-k)^2x}{(k+1)^2}$.

By letting $x = -\frac{(k+1)k}{(n-k)(n-k+1)}$, $a_{k-1} : a_k : a_{k+1}$ becomes $\frac{kn - k^2}{1-k^2 + nk + n} : -1 : \frac{kn - k^2}{1-k^2 + nk + n}$. So $a_k$ is greater than the neighboring terms in absolute value. An easy inspection now shows that the terms before and after $a_k$ form two alternating series, whose absolute values are decreasing. Therefore $\sum_{i < k} a_i$ and $\sum_{i>k} a_i$ are both less than $a_k\frac{kn - k^2}{1-k^2 + nk + n}$ in terms of absolute value.

We just have to refine our estimations a little bit. Indeed, the alternating series $\sum_{i>k}a_i$ can be grouped as $a_{k+1} + (a_{k+2} + a_{k+3}) + (a_{k+4}+a_{k+5}) +\dots$. We closely inspect this series

$(a_{k+2} + a_{k+3}) + (a_{k+4}+a_{k+5}) +\dots = a_{k+2} + (a_{k+3} + a_{k+4}) + (a_{k+5}+a_{k+6}) +\dots$,

But $|a_{k+2} + a_{k+3}| > |a_{k+3} + a_{k+4}|$, etc. We get $\left|\sum_{i > k+1} a_i\right| > \frac12 |a_{k+2}|$. This allows us to state

$$\left|\sum_{i>k} a_i\right| < |a_{k+1}| - \frac12 |a_{k+2}|$$

So $p\left(-\frac{(k+1)k}{(n-k)(n-k+1)}\right) = \mathrm{sgn}(a_k) \left(|a_k| - |a_{k+1}| - |a_{k-1}| + \frac12 |a_{k+2}| + \frac12 |a_{k-2}| \right)$. And it turns out that the stuff in the parentheses is positive. So It has the same sign as $a_k$.

The above choice for $x$ should (?) also work for $k=0,1,n-1,n$. So summing up, for each $k$, we have a choice of $x$ such that $p(x)$ has the same sign as $a_k$, which is just $(-1)^k$. And note that the $n+1$ choices of $x$ are increasing. Hence, we have $n$ alternation of signs of $p(x)$. So $p(x)$ has $n$ roots.

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  • $\begingroup$ Sadly, there is a major flaw: the estimation of $|a_{k+2}+a_{k+3}|$ is not correct, sometimes $|\sum_{i>k+1}a_i| < \frac12|a_{k+2}|$... $\endgroup$
    – Trebor
    Jul 9, 2021 at 9:32

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