The following result trivialize the problem (by using induction on $n$): F. F. Bonsall and Morris Marden, Zeros of self-inversive polynomials, Proc. Amer. Math. Soc. vol. 3 (1952) pp. 471-475. wrote: Let $g(z)=b_0+b_1z+b_2z^2+...+b_mz^m\in \mathbb{C}[x]$ be a self-inversive polynomial; that is, let $$b_i=u\overline{b_{m-i}}$$for all $i=0,1,2,...,m$ while $|u|=1$. Then $g(z)$ and its derivative $g'(z)$ have same number of zeros in the closed exterior of the unit circle $C$.

But there surely must be a simpler proof?

We don't need the full strength of that theorem. First note that when $\lambda = 0$, $f$ has $n$ distinct roots on the unit circle. Now let $\lambda$ increase. The only way the result can fail is if $f$ has a double root (this can be proved by writing $f$ as a function of the form $g(x+\frac{1}{x})$). But the derivative of $f_n(x)$ is just a multiple of $f_{n-1}(\frac{x}{\lambda})$. So by induction a double root cannot occur.

nevermind

Well, I guess you need to first explain where you think my argument is false :p

angiland wrote: Now let $\lambda$ increase. The only way the result can fail is if $f$ has a double root (this can be proved by writing $f$ as a function of the form $g(x+\frac{1}{x})$) Can you please give more details ?

How does the result from post #2 help if assumption there is a conclusion we have to proof?

Can you post "2018 China TST 4 Day 2 Problem 4"?

N.T.TUAN wrote: Can you post "2018 China TST 4 Day 2 Problem 4"? It is said that Problem 4 has some subtle association with the 2017 ISL. So they blocked the topic. Perhaps it can be post only after this year's IMO.

angiland wrote: We don't need the full strength of that theorem. First note that when $\lambda = 0$, $f$ has $n$ distinct roots on the unit circle. Now let $\lambda$ increase. The only way the result can fail is if $f$ has a double root (this can be proved by writing $f$ as a function of the form $g(x+\frac{1}{x})$). But the derivative of $f_n(x)$ is just a multiple of $f_{n-1}(\frac{x}{\lambda})$. So by induction a double root cannot occur. I really didn't get this argument (in red), but anyway, as written it's not valid, because $f$ also doesn't have a double root when $\lambda>0$. So, in the same way we would manage to prove the statement when $\lambda>0$, which is false.

Photaesthesia wrote: Suppose the real number $\lambda \in \left( 0,1\right),$ and let $n$ be a positive integer. Prove that the modulus of all the roots of the polynomial $$f\left ( x \right )=\sum_{k=0}^{n}\binom{n}{k}\lambda^{k\left ( n-k \right )}x^{k}$$are $1.$ We prove it by induction on $n$. For the induction step we use only the following easily checked fact: $$(1)\,\,\,\,\,\,\,\,\,\, f'_n(z)=n\lambda^{n-1}f_{n-1}\left(\frac{z}{\lambda}\right)\,,\,z\in \mathbb{C} $$Note that we can write $$(2)\,\,\,\,\,\,\,\,\,\, f_n(z)=z^{n/2}g_n(z).$$where $g_n(z)\in \mathbb{R}$ when $|z|=1$. Let $z=e^{i\varphi}$. Then $dz=ie^{i\varphi}d\varphi$, hence $\frac{df_n(z)}{dz}=\frac{df_n(\varphi)}{d\varphi}(-i)e^{-i\varphi}$. Starting from $(2)$, we get: \begin{align*} f'_n(z)&= -ie^{-i\varphi} \left( e^{i\frac{n}{2}\varphi} g_n(e^{i\varphi})\right)_{\varphi}'\\ &=\frac{n}{2}e^{i(\frac{n}{2}-1)\varphi}g_n(\varphi)-ie^{i(\frac{n}{2}-1)\varphi}g'_n(\varphi); \end{align*}Here, for bravity, we write $g_n(\varphi)$ instead of $g_n(e^{i\varphi})$. Using $(1)$ it follows: $$\frac{n}{2}g_n(\varphi) -i g'_n(\varphi)= z^{-\left(\frac{n}{2}-1\right)}f_{n-1}\left(\frac{z}{\lambda} \right)$$where $z=e^{i\varphi}$. It follows: $$(3)\,\,\,\,\,\,\,\,\,\,g_n(\varphi)= \frac{2}{n}\mbox{ Re }\left(z^{-\left(\frac{n}{2}-1\right)}f_{n-1}\left(\frac{z}{\lambda} \right)\right)$$ Now, let us induct on $n$. For $n=1$, obviously $f_1(z)=z+1$ has one zero on $|z|=1$. Suppose it's true for $f_{n-1}(z)$. To prove it for $f_n$ it's enough to establish that the LHS of $(3)$ has $n$ zeroes when $\varphi\in \left[0,2\pi\right)$. Let $u=\frac{z}{\lambda}$. When $z$ runs around the unit circle, $u$ runs around a circle $C$ with radius $\frac{1}{\lambda}>1$. Assume now $n$ is even. According to the argument principle, the winding number of the RHS of $(3)$ equals $(N-P)$ , where $N$ is the number of zeroes of $h(u):=u^{-(\frac{n}{2}-1)} f_{n-1}(u)$ inside $C$ and $P$ is the number of poles of $h(u)$ inside $C$. Obviously $P=\frac{n}{2}-1$ (a pole at $z=0$ of order $\frac{n}{2}-1$) and $N=n-1$ (according to to the induction hypothesis $f_{n-1}$ has $n-1$ zeroes inside $C$. Here we use $\lambda<1$). Thus, the winding number of $h(u)$ around $C$ is $\frac{n}{2}$. It means when $u$ makes one circle around $C$, $h(u)$ winds $\frac{n}{2}$ times around the origin. Hence $\mbox{Re}\,h(u)$ hits the zero $n$ times, which proves $f_n(z)$ has $n$ zeroes on the unit circle. For $n$ - odd, we cannot apply exactly the same trick, since $z^{-1/2}$ is not meromorphic. But, in that case we can lay $\varphi=2\psi$ on the LHS of $(3)$, which is virtually laying $u=v^2$ in $h(u)$. Then, it's allowed to apply the same argument and notice that $\varphi=2\psi, \psi\in [0,2\pi)$ doubles the number of zeroes of the LHS of $(3)$.

Nice problem, it appears to be extremely hard in the contest. Here are two solutions, the second one is mine.

By, the way, this problem has a post at MathOverflow which is worth looking at for anyone who found this problem interesting. A user pointed out that this problem is a special case of the Lee-Yang Theorem: Lee-Yang Theorem wrote: Let $G$ be a finite graph with vertex set $V$ and edge set $E$. Let $\beta\in (0, 1)$ be a real number. For a "spin function" $\sigma: V\to \{-1, 1\}$, let $m(\sigma)$ be the number of vertices with positive spin and $d(\sigma)$ be the number of edges with vertices of opposite spin. Let \[Z_{\beta}(x)=\sum_{\sigma}\beta^{d(\sigma)}x^{m(\sigma)}\]where the sum is over all possible spin functions. Then all roots of $Z_{\beta}$ lie on the unit circle.

dgrozev wrote: angiland wrote: First note that when $\lambda = 0$, $f$ has $n$ distinct roots on the unit circle. Now let $\lambda$ increase. The only way the result can fail is if $f$ has a double root (this can be proved by writing $f$ as a function of the form $g(x+\frac{1}{x})$). But the derivative of $f_n(x)$ is just a multiple of $f_{n-1}(\frac{x}{\lambda})$. So by induction a double root cannot occur. I really didn't get this argument (in red), but anyway, as written it's not valid, because $f$ also doesn't have a double root when $\lambda>0$. So, in the same way we would manage to prove the statement when $\lambda>0$, which is false. A precise statement of the red argument could be: $A=A_n=\{\lambda|f_{n,\lambda}\text{ has }n\text{ distinct roots on the unit circle}\}$ is open in $\mathbb{R}$. For any $\lambda\in A$, take a disc around each root of $f_\lambda$ such that they don't intersect each other and are all stable under inversion against the unit circle. Then for any $\lambda'$ sufficiently close to $\lambda$, each disc still contains exactly one root of $f_{\lambda'}$, while $z$ is a root $\Rightarrow$ $\overline{z}^{-1}$ is also root $\Rightarrow$ $z=\overline{z}^{-1}, |z|=1$. In short, "on the unit circle" sounds like a closed condition but in our circumstance it's forced by $z=\overline{z}^{-1}$ and doesn't matter. Now we have $\overline{A}\subseteq\{\lambda|f_\lambda\text{ has all of its roots on the unit circle}\}$ and \[\partial A\subseteq\{\lambda|f_\lambda\text{ has all of its roots on the unit circle and there are multiple ones}\},\]and since $A$ intersects $(0,1)$ if $(0,1)\not\subset A$ then $\partial A$ intersects $(0,1)$. Also in the induction \begin{align*} (0,1)\subseteq A_{n-1}&\Rightarrow\forall\lambda\in\overline{A_n}\cap(0,1),f_n\text{ and }f_{n-1}(\frac{x}{\lambda})\text{ have no common roots}\\ &\Rightarrow\forall\lambda\in\overline{A_n}\cap(0,1),f_n\text{ has no multiple roots}\\ &\Rightarrow \partial A_n\cap(0,1)=\emptyset\Rightarrow (0,1)\subseteq A_n, \end{align*}in the first implication $|\lambda|\neq 1$ is used, so it doesn't generalise beyond $(0,1)$.

iceillusion wrote: ...Then for any $\lambda'$ sufficiently close to $\lambda$, each disc still contains exactly one root of $f_{\lambda'}$, while $z$ is a root $\Rightarrow$ $\overline{z}^{-1}$ is also root $\Rightarrow$ $z=\overline{z}^{-1}, |z|=1$ ... You cannot claim $|z|=1.$ Ok, I agree that if $z_1:=z$ is a root then $z_2:=\overline{z}^{-1}$ is also a root. But this does not imply $z_1=z_2.$ It just means $(z_1,z_2)$ is a pair of roots for which $z_2=\overline{z_1}^{-1}$ and $z_1=\overline{z_2}^{-1}$.

iceillusion wrote: For any $\lambda\in A$, take a disc around each root of $f_\lambda$ such that they don't intersect each other and are all stable under inversion against the unit circle. Then for any $\lambda'$ sufficiently close to $\lambda$, each disc still contains exactly one root of $f_{\lambda'}$, while $z$ is a root $\Rightarrow$ $\overline{z}^{-1}$ is also root $\Rightarrow$ $z=\overline{z}^{-1}, |z|=1$. dgrozev wrote: You cannot claim $|z|=1.$ Ok, I agree that if $z_1:=z$ is a root then $z_2:=\overline{z}^{-1}$ is also a root. But this does not imply $z_1=z_2.$ It just means $(z_1,z_2)$ is a pair of roots for which $z_2=\overline{z_1}^{-1}$ and $z_1=\overline{z_2}^{-1}$. Please have a look at my complete arguments (pretty simple, but it seems in your mind topology doesn’t kind of come into play) before proceeding. The root $z$ of $f_{\lambda’}$ is in a small neighbourhood of $f_\lambda$, which contains only one root of $f_{\lambda’}$ and is also symmetric under inversion w.r.t. the unit circle, hence the root $\overline{z}^{-1}$ is also in the neighbourhood and its has to equal $z$.

iceillusion wrote: ...it seems in your mind topology doesn’t kind of come into play... Yeah, but only in case it's very poorly explained! Anyway, I got it. Your argument in #18 is ok. Moreover, it deserves attention and I could have thanked had you been a little bit more polite! Just a little bit. So, I want to comment the idea, since I think this is the core truth about this problem and could be used in many other more general situations as for example in the mentioned in #16 Lee-Yang theorem. Suppose we have a family of polynomials $P_{\lambda}(z), \lambda\in\mathbb{R}$ (it could be further generalized considering a family of holomorphic functions over some index set $\Lambda$ which is a topological space). Suppose the following two conditions are satisfied (i) For any fixed $\lambda$, $P_{\lambda}(z)$ is continuous as a function of $\lambda$ for any fixed $z\in\mathbb{C}$ and it holds uniformly on $z\in K$ for any compact set $K\subset \mathbb{C}.$ In other words, a slight change of $\lambda$ does not disturb $P_{\lambda}(z)$ too much on any compact set. (ii) if $z$ is a root of $P_{\lambda}(z)$ then $1/z$ and $\overline{z}$ also are. Having satisfied these two prerequisites, the following claim is true. If for some $\lambda_0$ all roots of $P_{\lambda_0}(z)$ are simple and lie on the unit circle then the same holds for all $\lambda$ in some neighbourhood of $\lambda_0.$ The idea is simple. Arguing by contradiction, suppose it fails. Then for any $\lambda$ as close to $\lambda_0$ as we want, there exists a root $z$ of $P_{\lambda}$ with $|z|\ne 1$. But then $1/\overline{z}$ is also a root of $P_{\lambda}.$ Due to (i), it would be as close to $z$ as we want. But (again by (i)) it is only possible if $P_{\lambda_0}$ has a double root, which is contradiction. Back to our particular problem. (i) and (ii) obviously hold. Since for $\lambda=0$, the roots of $f_{\lambda}$ are simple and lie on $|z|=1$ the same is true for some neighbourhood of $0$ and we can "push" $\lambda$ to the right. By (i) the only thing which can prevent us for pushing $\lambda$ further to the right is if $P_{\lambda}$ has double roots. But fortunately, it cannot happen for $\lambda\in (0,1),$ hence all the roots of $P_{\lambda}$ are on the unit circle for any $\lambda\in (0,1).$

dgrozev wrote: Yeah, but only in case it's very poorly explained! Anyway, I got it. Your argument in #18 is ok. Moreover, it deserves attention and I could have thanked had you been a little bit more polite! Just a little bit. ... (ii) if $z$ is a root of $P_{\lambda}(z)$ then $1/z$ and $\overline{z}$ also are. First I like this slow-paced and intuitive exposition (a small improvement: in (ii) we just need that $1/\overline{z}$ is also a root), it certainly suits the "High School Olympiads" board better than #18, while the later aims to give a brief and fully rigorous account for people with sufficient background knowledge. When composing #18 I presume you master the basics of undergraduate mathematics (my apologies if I was wrong). In that case I doubt if the explanation itself is "very poor" or is just very poorly processed (when your intuition is actively working you realise that when $z$ approaches a fixed point on the unit circle so does $\overline{z}^{-1}$, so I wonder if any further explanation is really necessary). I believe in a math discussion board a basic courtesy is to make effort to make sense of other people’s math statements (so that people don’t need to type duplicated contents), especially before saying "it’s not valid" or "You cannot claim" (which implies that other people didn’t manage to realise certain mistakes, while it’s a basic courtesy in life not to assume so until you’re sure the mistakes really exist). Therefore it surprised me a bit when someone failing to do so commented on other’s politeness.

why is it that i get the answer to be n using vietas sum of roots theorem because the first term has the cofficient (1) and second to be +,-(n) sum is (second coefficient)/(first)

The problem looks like it has some recursive substructure, and that solves the problem: We can bash to see $f_n(x)=f_{n-1}(\lambda x)+x\lambda^{n-1} f_{n-1}(\frac{x}{\lambda})$ Proceed by induction; if $f_{n-1}$ has all roots of multiplicity 1, then its roots can be paired $(r, r^{-1})$ with the exception pf $r=-1$ when $n-1$ is odd. Now we bash moduli: $|f_{n-1}(\lambda x)|=x\lambda^{n-1} |f_{n-1}(x/ \lambda)|$ If $|x|>1$ then $ \prod |\lambda x-r|>\prod |x-\lambda r|$ However, notice $|\lambda x-r|<|x-\lambda r|$: the r plays a bigger factor in lhs. Squaring and Expanding gives $(\lambda^2-1)(|x|^2-1)$ is the difference of the square of their moduli.

angiland wrote: We don't need the full strength of that theorem. First note that when $\lambda = 0$, $f$ has $n$ distinct roots on the unit circle. Now let $\lambda$ increase. The only way the result can fail is if $f$ has a double root (this can be proved by writing $f$ as a function of the form $g(x+\frac{1}{x})$). But the derivative of $f_n(x)$ is just a multiple of $f_{n-1}(\frac{x}{\lambda})$. So by induction a double root cannot occur. Terrific intuition!!