Prove a polynomial has a nonreal root

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Prove a polynomial has a nonreal root

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Source: USAMO 2025/2

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#1 h Mar 20, 2025, 12:00 PM • 4 Y

Y by MathRook7817, NaturalSelection, LostDreams, NO_SQUARES

Let $n$ and $k$ be positive integers with $k<n$ . Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$ , the product $a_0a_1\cdots a_k$ is zero. Prove that $P(x)$ has a nonreal root.

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#2 h Mar 20, 2025, 12:02 PM • 3 Y

Y by MathRook7817, megarnie, LostDreams

Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$ . WLOG $k=n-1$ , as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$ .

By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$ . Since $r_i\neq 0$ for all $i$ , this coefficient is not the leading or constant coefficient, so it has $k-1=n-2$ choices. By pigeonhole, there exists $1\leq i<j\leq n$ such that $Q(x):=\frac{P(x)}{x-r_i}$ and $R(x):=\frac{P(x)}{x-r_j}$ have no $x^\alpha$ term, for some $\alpha\geq 1$ . WLOG $i=n-1$ and $j=n$ . Hence
$\begin{align*} P(x)&=(x-r_1)\cdots(x-r_{n-2})(x-r_n)\\ Q(x)&=(x-r_1)\cdots(x-r_{n-2})(x-r_{n-1}). \end{align*}$ Then $P(x)-Q(x)=(r_{n-1}-r_n)(x-r_1)\cdots(x-r_{n-2})$ has no $x^\alpha$ term, so $A(x):=(x-r_1)\cdots(x-r_{n-2})$ has no $x^\alpha$ term since $r_{n-1}\neq r_n$ . Also, $r_{n-1}P(x)-r_nQ(x)=x(r_{n-1}-r_n)(x-r_1)\cdots(x-r_{n-2})$ has no $x^\alpha$ term so $A(x)$ has no $x^{\alpha-1}$ term.

Claim. $A(x)$ cannot have $n-2$ distinct real roots.
Proof. Suppose FTSOC $A(x)$ has $n-2$ distinct real roots. We claim that $A^{(m)}(x)$ , the $m$ th derivative of $A(x)$ , has $n-2-m$ distinct real roots for all $m\leq n-2$ . We proceed by induction on $m$ with the base case $m=0$ trivial.

Assume $A^{(m)}(x)$ has $n-2-m$ distinct real roots. Between every $2$ consecutive real roots there exists a local extremum, where $A^{(m+1)}(x)$ is $0$ . $A^{(m)}(x)$ has $n-2-m-1$ pairs of consecutive real roots so $A^{(m+1)}(x)$ has $n-2-(m+1)$ distinct real roots, completing the induction step.

$A^{(\alpha-1)}$ has no $x^{(\alpha-1)-(\alpha-1)}=x^0$ or $x^{\alpha-(\alpha-1)}=x$ term, so $0$ is a double root of $A^{(\alpha-1)}$ . However, all the roots of $A^{(\alpha-1)}$ are distinct and real, a contradiction. $\square$

This is a contradiction since $r_1,\,\ldots,\,r_{n-2}$ are distinct real numbers. Thus $P$ has a nonreal root. $\square$

This post has been edited 1 time. Last edited by KevinYang2.71, Mar 21, 2025, 12:12 AM

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#3 h Mar 20, 2025, 12:10 PM

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tfw you wrote up a fakesolve that turned out to be correct

KevinYang2.71 wrote:

Claim. $A(x)$ cannot have $n-2$ distinct real roots.

how many points for claiming this but not proving it

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#4 h Mar 20, 2025, 12:11 PM • 1 Y

Y by KevinYang2.71

Assume for contradiction that all roots are real. Note that any divisor of $P$ with degree $>k$ will still satisfy the problem condition, hence we may WLOG assume $\deg P = k+1$ .

Denote the $d$ -th elementary symmetric sum of a set $S$ as $p_d(S)$ .

Reformulation of problem by Vieta wrote:

Distinct nonzero real numbers $r_1, \dots , r_{k+1}$ satisfy the following property: for each subset $S \subseteq \{r_1, \dots , r_{k+1}\}$ of size $|S|=k$ , there exists some $1 \le d_S < k$ such that $p_{d_S}(S) = 0$ .

Show this is impossible.

Since there are $k+1$ such subsets, by pigeonhole there are two distinct subsets $S_1$ and $S_2$ with $d_{S_1} = d_{S_2} \equiv d$ . WLOG let $S_1 = \{r_1, \dots , r_k\}$ and $S_2 = \{r_2, \dots , r_{k+1}\}$ , and denote $S_3 \equiv S_1 \cap S_2 = \{r_2, \dots , r_k\}$ ; then
$p_d (S_1) = p_d (S_2) \implies (r_{k+1}-r_1)p_{d-1}(S_3) = 0 \implies p_{d-1}(S_3) =0$ since $r_1 \neq r_{k+1}$ by assumption. Furthermore, we have
$0 = p_d(S_1) = r_1 p_{d-1}(S_3) + p_d(S_3) \implies p_d(S_3) = 0.$ But it turns out to be impossible to have $p_d(S_3) = p_{d-1}(S_3) = 0$ :

Lemma: A set of distinct nonzero real numbers cannot have two consecutive elementary symmetric sums both equal to zero. Equivalently, a polynomial with distinct nonzero roots cannot have two consecutive zero coefficients.

Proof: The second sentence follows by Descartes' rule of signs.

This post has been edited 2 times. Last edited by VulcanForge, Mar 20, 2025, 12:15 PM

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#6 h Mar 20, 2025, 12:19 PM

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scannose wrote:

tfw you wrote up a fakesolve that turned out to be correct

KevinYang2.71 wrote:

Claim. $A(x)$ cannot have $n-2$ distinct real roots.

how many points for claiming this but not proving it

2 for everything before using derivative maybe

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#7 h Mar 20, 2025, 1:07 PM

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This exists https://artofproblemsolving.com/community/c6h3486913_a_polynomial_problem

it's not the same problem, but it is essentially the main claim

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#8 h Mar 20, 2025, 1:09 PM • 1 Y

Y by megarnie

Assume for the sake of contradiction that $P$ has $n$ real roots, say $r_1, r_2, \dots, r_n.$ If $n > k+1,$ we just need to consider a subset of these roots in the $n=k+1$ case to get our contradiction. Thus assume $n = k+1.$

Now for a subset $U \subset \{1,2,\dots,n\},$ we define $S_{k,U}$ to be the degree $k$ symmetric sum of the $r_i$ for all $i \in U.$ Also let $U_i = \{1,2,\dots, n\} \setminus \{i\}$ Then by Vieta's Formulas, we are given that for all $1 \le i \le k+1,$
$\begin{align*} S_{1,U_1} \cdot S_{2,U_1} \cdots S_{k,U_1} &=0, \\ S_{1,U_2} \cdot S_{2,U_2} \cdots S_{k,U_2} &=0, \\ \vdots \\ S_{1, U_{k+1}} \cdot S_{2,U_{k+1}} \cdots S_{k, U_{k+1}} &= 0. \end{align*}$
By PHP two of the numbers in the same column are equal to $0.$ Suppose WLOG that $S_{i, U_1} = S_{i, U_2} = 0.$ Some algebraic manipulation gives $S_{i-1, \{3,4,\dots,k+1\}} = S_{i, \{3,4,\dots,k+1\}} = 0.$ This means that the polynomial $Q(x) = \frac{P(x)}{(x-r_1)(x-r_2)}$ has two consecutive coefficients equal to $0.$

Claim: This cannot happen.
Proof: By Rolle's Theorem, the derivative $Q'(x)$ has $n-1$ distinct real roots as well, along with two consecutive coefficients which are equal to $0.$ We keep taking the derivative until we end up with something divisible by $x^2,$ contradiction as there can be no double roots.

Thus we have established a contradiction, hence done.

how many points would I get up to before the claim?

This post has been edited 3 times. Last edited by EpicBird08, Apr 2, 2025, 8:39 PM

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#9 h Mar 20, 2025, 1:15 PM • 1 Y

Y by Mintylemon66

The following claim simplifies our problem a bit:
Claim:
It suffices to prove the statement for $n = k + 1$
Proof:
Assume the statement is true for $n = k + 1.$
Take a subset of $k + 1$ roots of $P.$ The set of degree $k$ polynomials generated by these $k + 1$ roots is a subset of the set of degree $k$ polynomials generated by all $n$ roots.

Thus, if we apply our hypothesis on this subset, we are get at least one of these roots is complex.
Now, we define a few things:
Assume $\ell \leq m.$ Say an equation is \textbf{\emph{type} $\ell$ of set $\{r_1, \dots, r_m\}$ } if the equation has form $\sum_{\stackrel{S\subseteq \{r_1, \dots, r_m\}}{|S| = \ell}}\prod_{r\in S}r = 0$ In other words, it is a cyclic sum of degree $\ell$ polynomials. We will call $\ell$ the degree. We will also say it is missing $r_j$ if $r_j \not\in \{r_1, \dots, r_m\}$
For example, $r_1r_2 + r_1r_3 + r_2r_3$ is type $3$ of $\{r_1, r_2, r_3\}.$
Claim:
If we prove that there can't be two equations of the same degree, we are done.
Proof:
First, note that we have $\binom{k + 1}{k} = k + 1$ equations. However, there are only $k$ possible degrees, which are $1, \dots, k$ (note that $k$ can't work, but this doesn't really matter). By PHP, this implies that there can't be two equations of the same degree.
\end{proof}
Claim:
There can't be two equations of the same degree
Proof:
WLOG, we assume $r_1, r_2$ are missing from the first and second equations, respectively. Also assume the degree is $0 < \ell < k.$ This gives:
\begin{equation}
r_2r_3\dots r_{\ell + 1} + r_2r_3\dots r_{\ell + 2} + \dots + r_{n - \ell + 1}\dots r_{n} = 0
\end{equation}
\begin{equation}
r_1r_3\dots r_{\ell + 1} + r_1r_3\dots r_{\ell + 2} + \dots + r_{n - \ell + 1}\dots r_{n} = 0
\end{equation}
From simple cancellation of comparing the two, we have that the type $\ell - 1$ equation of $\{r_3, \dots, r_{n}\}$ must be 0(since otherwise, $r_1 = r_2$ ):
\begin{equation}
r_3r_4\dots r_{\ell + 1} + r_3r_4\dots r_{\ell + 2} + \dots + r_{n - \ell + 1}\dots r_{n} = 0
\end{equation}
If we took $(1) - r_1\cdot (3),$ we get that the type $\ell$ equation of $\{r_3, \dots, r_n\}$ is 0, or:
\begin{equation}
r_3r_4\dots r_{\ell} + r_3r_4\dots r_{\ell + 1} + \dots + r_{n - \ell + 2}\dots r_{n} = 0
\end{equation}
This implies that there is a polynomial dividing $P$ that has two consecutive zero coefficients. Observe that by Rolle's Theorem, its derivative also has real roots. By simply inducting down on $k,$ we are done. The base case is obvious.

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#10 h Mar 20, 2025, 1:21 PM • 1 Y

Y by ihatemath123

This link also shows that a polynomial with distinct real roots can’t have $3$ consecutive coefficients in a geometric series. This could be another way to show that $2$ consecutive coefficients can’t be $0$ , but I think the normal approach is easier and more direct.

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#11 h Mar 20, 2025, 1:36 PM

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How many points for falling into fakesolve trap (I assume 0 cuz its just special case)

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#12 h Mar 20, 2025, 2:15 PM

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how many points for saying proof by contradiction, none of the roots can be zero cause the constant is zero, and writing out all the polylomials in forms, stating that its probably descartes rules of signs or vietas and ending it from there

This post has been edited 1 time. Last edited by KadenC2026, Mar 20, 2025, 2:18 PM
Reason: To clarify post

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#13 h Mar 20, 2025, 2:19 PM

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i don't think u can just assume k = n-1. My proof used the fact that elementary symmetric sums are multilinear, so you can't have the same elementary symmetric sum (i.e. both the sum of the roots, both the product, or both the sum of pairwise products) of 2 sets of k roots with only 1 root in a given set but not the other be both 0.

Then I had to prove that max_2 (n,k) is less than (n choose k) / k-1. Is that true? Here max_2 (n,k) is the maximum possible size of a collection of k-element subsets of an n-element set such that no 2 subsets intersect in k-1 or k elements.

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#14 h Mar 20, 2025, 2:20 PM • 1 Y

Y by scannose

k=n-1 is sufficient because otherwise you may take a polynomial Q(x) of degree k+1, with A|Q|P, and the pair of k, k+1 is done by above

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#17 h Mar 20, 2025, 2:23 PM

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awesomeguy856 wrote:

k=n-1 is sufficient because otherwise you may take a polynomial Q(x) of degree k+1, with A|Q|P, and the pair of k, k+1 is done by above

oh no! i could have made this much easier! does my solution with the multilinear polynomials and max_2 work?

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#18 h Mar 20, 2025, 2:28 PM

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Does anyone know how many “mohs” this question is, whatever that means?

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#19 h Mar 20, 2025, 2:30 PM

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cushie27 wrote:

Does anyone know how many “mohs” this question is, whatever that means?

Math Olympiad Hardness Scale

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OronSH

#20 h Mar 20, 2025, 2:30 PM • 1 Y

Y by Pengu14

cushie27 wrote:

Does anyone know how many “mohs” this question is, whatever that means?

ive heard 30

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#21 h Mar 20, 2025, 2:32 PM • 6 Y

Y by OronSH, mineric, ihatemath123, peace09, scannose, trk08

ts pmo $\text{[math]}$

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#22 h Mar 20, 2025, 2:32 PM • 1 Y

Y by OronSH

OronSH wrote:

cushie27 wrote:

Does anyone know how many “mohs” this question is, whatever that means?

ive heard 30

I hope that means bronze medal cutoff shouldn’t be affected too much by this problem…

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#23 h Mar 20, 2025, 2:32 PM • 1 Y

Y by OronSH

Observe we only need to prove for $k = n-1$ ; indeed, $P$ either has no real roots, or it has a real root $r$ , and we can induct downwards by considering $P(x)/(x-r)$ until $k = n-1$ .

Let $\sigma_i(x_1, x_2, \dots, x_\alpha)$ denote the $i$ th elementary symmetric sum in the $\alpha$ varaibles $x_1$ , $\dots$ , $x_\alpha$ . Recall the identity $\sigma_i(x_1, x_2, \dots, x_n, y) = \sigma_i(x_1, x_2, \dots, x_n) + y\sigma_{i-1}(x_1, x_2, \dots, x_n).$
Now suppose FTSOC that $P$ has all real roots. Then as $\binom{n}{n-1} = n > n-1$ , there exist two subsets $S \subset\{1, 2, \dots, n\}$ with $|S| = n-1$ such that the corresponding coefficient of $\prod(x-r_i)$ which is $0$ is the same. In other words, we can order the roots of $P$ in such a way that for some $j$ ,
$\sigma_j(r_1, r_2, \dots, r_{n-2}, r_{n-1}) = \sigma_j(r_1, r_2, \dots, r_{n-2}, r_n) = 0.$ Using the identity above, and assuming all roots are distinct, we obtain that $\sigma_{i-1}(r_1, \dots, r_{n-2}) = 0$ , and similarly $\sigma_i(r_1, \dots, r_{n-2}) = 0$ .

By considering the polynomial $\prod_{i=1}^{n-2} (x-r_i),$ we are done by Descartes' rule of signs (a quick combinatorial argument shows that in fact we can only get up to $n-4$ real roots.)

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#24 h Mar 20, 2025, 2:34 PM

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Does anyone know how day 1 was compared to previous years contests? I'm a first timer and I'm trying to see if I have a chance at mop.

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#25 h Mar 20, 2025, 2:44 PM

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If I got everything in Kevin's (#2) solution before claiming $A(x)$ cannot have $n-2$ distinct real roots would that be $2$ partials?

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#26 h Mar 20, 2025, 2:52 PM

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Suppose $P(x)$ had no nonreal roots. We can assume $P(x)$ has degree $k + 1$ , as we can always find a polynomial $Q(x) \mid P(x)$ such that $Q(x)$ has no nonreal roots. We can also assume $P(x)$ is monic by scaling. Also, the $k = 1$ case is trivial as the constant term must be nonzero, so fix $k \geq 2$ . Let the roots of $P(x)$ be $r_1, r_2, \dots, r_{k+1}$ .

Now consider the degree $k-1$ polynomials $Q_c(x) = \prod_{\stackrel{i=0}{i\neq c}}^k (x - r_i)$ for integers $1 \leq c \leq k$ . If $Q_c(x) = x^{k-1} + \alpha_{k-2}x^{k-2} + \cdots + \alpha_0$ , then for $Q_c(x)(x - r_{k+1})$ to have some coefficient equal to $0$ , expanding shows that we must have $r_{k+1} \in \left\{\alpha_{k-2}, \frac{\alpha_{k-3}}{\alpha_{k-2}}, \dots, \frac{\alpha_0}{\alpha_1}\right\}.$
There are at most $k - 1$ values in this set. Fix the value of $r_{k+1}$ . Now consider the degree $k - 2$ polynomials $Q_{c,d}(x) = \gcd(Q_c(x), Q_d(x)) = \prod_{\stackrel{i=0}{i\neq c,d}}^k (x - r_i)$ . If $Q_{c,d}(x) = x^{k-2} + b_{k-3}x^{k-3} + \cdots + b_0$ and $Q_{c,d}(x)(x-s) = x^{k-1} + B_{k-2}x^{k-2} + \cdots + B_0$ , then for any fixed $i$ , $\frac{B_i}{B_{i+1}} = r_{k+1}$ is a linear equation in $s$ .

Case 1: For all $i$ , $\frac{B_i}{B_{i+1}} = r_{k+1}$ has at most one solution for $s$ . Then, this means that $Q_{c,d}(x)(x - r_c) = Q_d(x)$ and $Q_{c,d}(x)(x - r_d) = Q_c(x)$ must have different $i$ (if one exists) such that $\frac{B_i}{B_{i+1}} = r_{k+1}$ . But there are only $k - 1$ such indices of $i$ and $k$ polynomials, so there exists some $c$ such that $Q_c(x)(x - r_{k+1})$ has no coefficient equal to $0$ , done.

This is the extent to what I got to in contest.

Case 2: There exists some $i$ such that $\frac{B_i}{B_{i+1}} = r_{k+1}$ has infinite solutions in $s$ . We show this case is impossible since it will result in a polynomial not having all distinct real roots.

Since $B_i = b_{i-1} - sb_i$ and $B_{i+1} = b_i - sb_{i+1}$ , rearranging gives $(r_{k+1}b_{i+1} - b_i)s = (r_{k+1}b_i - b_{i-1})$ . For this to have infinitely many solutions, we must have $r_{k+1}b_{i+1} - b_i = r_{k+1}b_i - b_{i-1} = 0$ , thus $b_{i+1}, b_i, b_{i-1}$ forms a geometric series of ratio $r_{k+1}$ .

Thus $Q_{c,d}(x)$ has three consecutive coefficients in geometric progression. Let $R(x) = (x - r_{k+1})Q_{c,d}(x)$ . It is easy to see that this polynomial has two consecutive coefficients equal to zero. If we show $Q_{c,d}(x)$ does not have all distinct real roots, then we are done.

Suppose $R(x)$ has all distinct real roots (with the exception of a possible root at $x = r_{k+1}$ with multiplicity $2$ ). Then between each root (or possibly at $x = r_{k+1}$ ) is a local extremum, so $R^\prime(x)$ also has all distinct real roots (with no exception at $x = r_{k+1}$ ), but with one lower degree. By the power rule, this polynomial also has two consecutive coefficients equal to zero, but with one lower degree. By doing this repeatedly, we eventually end up with a polynomial where the two consecutive coefficients have degree $1$ and $0$ respectively. But this polynomial has a double root at $x = 0$ , contradiction as $r_{k+1} \neq 0$ .

This post has been edited 4 times. Last edited by plang2008, Mar 31, 2025, 3:39 PM

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#27 h Mar 20, 2025, 2:56 PM

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if I did everything else, except forgetting to state Pigeonhole and k + 1, how many partials would that be

VulcanForge wrote:

Since there are $k+1$ such subsets, by pigeonhole there are two distinct subsets $S_1$ and $S_2$ with $d_{S_1} = d_{S_2} \equiv d$ . WLOG let $S_1 = \{r_1, \dots , r_k\}$ and $S_2 = \{r_2, \dots , r_{k+1}\}$ , and denote $S_3 \equiv S_1 \cap S_2 = \{r_2, \dots , r_k\}$ ; then

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#28 h Mar 20, 2025, 3:00 PM

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it's funny how half of the posts on this thread asked about partials (ok half is an exaggeration but)
this year seems harder than the last

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#29 h Mar 20, 2025, 3:01 PM

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scannose wrote:

it's funny how half of the posts on this thread asked about partials
this year seems harder than the last

I disagree, I thought p2 on 2024 amo was quite a bit harder

however this could also be because im biased and not great at combo

@below p3 apparently had more solves this year

difficulty in general depends a lot on your subject preference

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#30 h Mar 20, 2025, 3:02 PM

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megarnie wrote:

I disagree, I thought p2 on 2024 amo was quite a bit harder

however this could also be because im biased and not great at combo

even if p2 wasn't harder, p3 was

edit: oh womp womp

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deduck

#31 h Mar 20, 2025, 11:05 PM • 1 Y

Y by KevinYang2.71

i thought this was easier than p3 but maybe that's because i was being an idiot on the other one and scared of it

just stare at problem until u get two polynomials with $n$ roots from a set of $n+1$ distinct roots that have the same coefficient equal to $0$ , note that either the two sets of $n$ roots are the same, which is false, or the $x$ th and $x+1$ th symmetric sum involving $n-1$ roots that they share in common is both $0$ . However this implies that there's $2$ consecutive coefficients equal to $0$ when u write it the standard way, but it's obviously false it can happen if all roots are distinct and real

this is my last olympiad problem , i don't know what made me waste the better part of my childhood wasting my time on these while people more gifted or cheaters were going to beat me in the end so i didnt even get to do a single olympiad problem in olympiad

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#32 h Mar 20, 2025, 11:31 PM

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wait wtf i got to the two consecutive coefficients are zero and wrote “now i don’t know what to do” even though i just needed a literal fifteen words saying “we repeatedly differentiate until we get a double root, which is impossible by Rolle’s Theorem.”

grrrrrrrrr how many points do you think i get

if i miss mop b/c of this i’ll be sad

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#33 h Mar 20, 2025, 11:39 PM

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@above 0 $~$

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#34 h Mar 21, 2025, 1:02 AM

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solasky wrote:

wait wtf i got to the two consecutive coefficients are zero and wrote “now i don’t know what to do” even though i just needed a literal fifteen words saying “we repeatedly differentiate until we get a double root, which is impossible by Rolle’s Theorem.”

grrrrrrrrr how many points do you think i get

if i miss mop b/c of this i’ll be sad

im in a similar situation as you oops
some orz people on discord said it may be worth 2 points

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#35 h Mar 21, 2025, 1:04 AM

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I have the opposite issue compared to many. I proved the 2 consecutive coefficients part, but I failed the pigeonhole part since I didn't realize I could WLOG $k=n-1$ . I ended up writing there were at least $\left\lceil\frac{\binom{n}{k}}{k-1}\right\rceil$ subsets of $k$ roots sharing a $0$ coefficient by Pigeonhole, then trying to show two of these subsets shared $k-1$ roots, thus implying the two consecutive zero coefficients (which failed miserably). Anyone know how many points this would be?

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#36 h Mar 21, 2025, 1:29 AM • 2 Y

Y by Ilikeminecraft, Math4Life2020

Here, we have a solution without WLOG $k=n-1$ , without Descartes rule of signs, without Rolle, but is somehow not terrible?

Let $e_k$ denote the degree $k$ elementary symmetric polynomial.

Suppose for the sake of contradiction that all roots of $P(x)$ were real. Let the roots be $r_1,r_2,\dotsc,r_n$ . Let $R$ be the set of roots. Then for every subset $S$ of $k$ roots, $e_i(S)=0$ for some $1\le i\le k-1$ .

Define two subsets of $k$ elements to be adjacent if they share $k-1$ elements.
Lemma. If $\mathcal F$ is a family of $k$ -element subsets of $R$ such that no two are adjacent, then $|\mathcal F|\le\frac{1}{k+1}\binom nk$ .
Proof. Every set $T$ in $\mathcal F$ is adjacent to exactly $k(n-k)$ sets outside of $\mathcal F$ because we may pick any element of $T$ to remove and any root not in $T$ to add. Meanwhile, any set $U$ not in $\mathcal F$ is adjacent to at most $n-k$ sets in $\mathcal F$ because when we remove an element of $U$ and add a root not in $U$ , the additional roots for the adjacent sets in $\mathcal F$ must be distinct. Counting then gives the desired result.

Now define $\mathcal F_i$ to be the family of $k$ -element subsets of roots such that $e_i$ vanishes. Then there are $k-1$ such families, and every size $k$ subset of the roots is in at least one family, so one family must have size at least $\frac{1}{k-1}\binom nk$ . Therefore, at least one family must contain two adjacent subsets. This implies that there are roots $r_1,r_2,\dotsc,r_k,r_k'$ such that
$e_i(r_1,r_2,\dotsc,r_k)=e_i(r_1,r_2,\dotsc,r_{k-1},r_k')=0.$ Now we expand these by writing
$e_i(r_1,r_2,\dotsc,r_{k-1})+r_ke_{i-1}(r_1,r_2,\dotsc,r_{k-1})=e_i(r_1,r_2,\dotsc,r_{k-1})+r_k'e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0,$ and since $r_k\neq r_k'$ , we have
$e_i(r_1,r_2,\dotsc,r_{k-1})=e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0.$
Suppose that $\{r_1,r_2,\dotsc,r_{k-1}\}$ is a set of real numbers such that $e_i(r_1,r_2,\dotsc,r_{k-1})=e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0$ for some $i$ , but no strict subset has this property. It suffices to show that such a set cannot exist.

Now we perform a bit of further expansion
$e_i(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-1}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=0,$ from which we may deduce that
$e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2.$
By minimality, none of $e_i$ , $e_{i-2}$ , or $e_{i-1}$ can vanish, for then two consecutive elementary symmetric polynomials must vanish. By "On the nonnegativity of generalized discriminants of quadratics",
$e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2\ge \frac{i(k-i)}{(i-1)(k-i-1)}e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2}).$ Since $e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2$ is strictly positive, we have a contradiction.

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#37 h Mar 21, 2025, 5:34 AM • 2 Y

Y by KevinYang2.71, Pengu14

This problem is kinda funny, I guess giving a rating is hard.
So FTSOC $P$ has $n$ real roots and WLOG it is monic and WLOG $k=n-1$ by simply shifting with the other monomials then define $P(x)$ as $(x-r_1) \cdots (x-r_n)$ then it happens that all of $\frac{P(x)}{x-r_i}$ for $1 \le i \le n$ happen to have one coefficient equal to $0$ , since there is at most $n-2$ choices for these so by pigeonhole there exists $i \ne j$ such that $\frac{P(x)}{x-r_i}$ and $\frac{P(x)}{x-r_j}$ are missing the same coefficient $x^{\ell}$ for $\ell \ge 1$ (this is because none of the roots are equal and none can be zero), let these be called $A(x), B(x)$ respectively and let $i,j$ be $n,n-1$ (it doesn't matter anyway lmao), then we have that $A(x)=(x-r_1) \cdots (x-r_{n-1})$ and $B(x)=(x-r_1) \cdots (x-r_{n-2})(x-r_n)$ which the means that $A(x)-B(x)=(r_n-r_{n-1})(x-r_1) \cdots (x-r_{n-2})$ is missing the $x^{\ell}$ coefficient but also notice that $r_nA(x)-r_{n-1}B(x)=(r_n-r_{n-1})x(x-r_1) \cdots (x-r_{n-2})$ is missing the $x^{\ell}$ coefficient but since $r_n \ne r_{n-1}$ then we do in fact have that $A(x)-B(x)$ is missing both the $x^{\ell}$ and $x^{\ell-1}$ coefficient. (Now we can wlog say $\ell$ is minimal).
Call it $C(x)$ then it has $n-2$ real roots, is $\deg C=n-2$ but also it has two consecutive missing coefficients so write it as $x^2D(x)+E(x)$ (essy to check that if $n=2$ this is trivial and if $n=3$ then you have cuadratics of the form $x^2+b$ which shows pairwise sum of roots is zero and thus all are zero, a contradiction!), so for $n \ge 4$ notice that $n-4=\deg D$ and that you can set it so that least degree term on $D$ is greater than $\deg E$ .
Now from derivative analytical definition and propeties we can check graphically that counting multiplicity a polynomial has all its roots real if and only if the derivative does as well, so now take the $\deg E+1$ -th derivative of $Q$ , then $Q^{(\deg E+1)}(0)=0$ but also $Q^{(\deg E+2)}(0)=0$ and these polynomials must have all of their roots real by chaining the fact that $Q$ does follow this so because of minimality of $\ell$ , it happens that $E$ has a non-zero leading coefficient and thus if it were the zero polynomial we are done by the same argument as if it weren't which we will see now.
So let $R(x)=Q^{(\deg E)}(x)=c(x-q_1) \cdots (x-q_m)$ , now focus on the product of monomials and call it $R_1(x)$ , then it happens that $\frac{R_1'(x)}{R_1(x)}=\sum_{i=1}^{m} \frac{1}{x-q_i}$ and thus by taking derivative w.r.t. $x$ it happens that $\frac{R_1'(x)^2-R_1''(x)R_1(x)}{R_1(x)^2}=-\sum_{i=1}^{m} \frac{1}{(x-q_i)^2}$ for all $x \ne q_j$ for $1 \le j \le m$ , however clearly none of the roots are zero once again but replacing $x=0$ we have LHS is zero while RHS isn't, contradiction!.
Therefore $P$ must have a non-real root thus we are done :cool:

:cool:

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#38 h Mar 21, 2025, 6:54 PM

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How many points for proving it suffices to prove k=n-1? I made a mistake for the part where I prove k=n-1 works, so I don’t think I’m getting any credit there.

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#39 h Mar 21, 2025, 7:17 PM

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This problem is certainly based on the following lemma:
Lemma: A polynomial with only real distinct roots can't have two consecutive zero coefficients
Proof: Let the polynomial be $P$ and the coefficients be for $x^k,x^{k+1}$ . Then $P^{(k)}$ is a polynomial which must have only real distinct roots, but it is divisible by $x^2$ , a contradiction.

Now to the problem. We consider the contrary. Take any $k+1$ of this $n$ roots, say they are $r_1,\ldots,r_{k+1}$ . Consider the polynomials of the form $\frac{(x-r_1)\ldots (x-r_{k+1})}{x-r_i}$ . $P$ is divisible by it, hence it has a zero coefficient. But its coefficients are elementary symmetric polynomials in variables $r_1,r_2,\ldots,r_{k+1}$ . Also, it is not the product since none of the roots is zero. So there are only $k-1$ options for the size of the elementary polynomial, and $k+1$ polynomials, by Pigeonhole there is a repetition. So $\sigma_i(r_1,\ldots,r_{k-1},r_k)=\sigma_i(r_1,\ldots,r_{k-1},r_{k+1})=0$ . But $\sigma_i(r_1,\ldots,r_{k-1},x)=x\sigma_{i-1}(r_1,\ldots,r_{k-1})+\sigma_i(r_1,\ldots,r_{k-1})$ , and as it is a linear function with two distinct roots, we must have $\sigma_{i-1}(r_1,\ldots,r_{k-1})=\sigma_i(r_1,\ldots,r_{k-1})=0$ . This contradicts the Lemma. Thus, $P$ must have a nonreal root.

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#40 h Mar 21, 2025, 10:19 PM

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How much do I get if I proved everything but assumed n=k+1 without justifying? I was a bit confused by the wording

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awesomeming327.

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#41 h Mar 22, 2025, 2:58 AM

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Assume on the contrary. We prove the result for $k=n-1$ . This is equivalent to the original problem because we can simply remove some of the roots of $P$ to reduce to this new problem anyway.

Let $P(x)=(x-r_1)(x-r_2)\dots(x-r_n)$ for some distinct nonzero roots $r_1$ , $r_2$ , $\dots$ , $r_n$ . Let $P_i(x)$ be $(x-r_1)\dots(x-r_{i-1})(x-r_{i+1})\dots(x-r_n)$ , which must have a zero coefficient. Note that there are $n$ polynomials $P_1$ , $P_2$ , $\dots$ , $P_n$ and $n-2$ coefficients at which this zero coefficient can appear. Therefore, there exists two polynomials (WLOG let them be $P_n$ and $P_{n-1}$ ) such that $P_n$ and $P_{n-1}$ both have a zero as an $x^i$ coefficient.

Let $(x-r_1)(x-r_2)\dots(x-r_{n-2})$ be $Q(x)$ . We have
$\begin{align*}P_n-P_{n-1} &= (x-r_{n-1})Q(x)-(x-r_n)Q(x)=(r_n-r_{n-1})Q(x) \\ r_nP_n-r_{n-1}P_{n-1} &= (r_nx-r_{n-1}r_n)Q(x)-(r_{n-1}x-r_{n-1}r_n)Q(x)=(r_n-r_{n-1})xQ(x)\end{align*}$ both of which have zero as an $x^i$ coefficient, which means that $Q(x)$ has a zero in both its $x^i$ and $x^{i-1}$ coefficients. We now prove the following claim, which finishes the problem:

Claim 1: Let $Q(x)=(x-r_1)(x-r_2)\dots(x-r_n)$ be a polynomial with distinct nonzero real roots. Then $Q(x)$ may not have two consecutive zero coefficients.

Let $R(x)$ be a polynomial with $r$ distinct roots. Then, $R'(x)$ has $r-1$ distinct roots because between each two roots of $R(x)$ , there will be a relative extrema. Therefore, if we keep taking derivatives of $Q$ , there will always be $\operatorname{deg} Q$ distinct roots. But since there are two consecutive zero coefficients, eventually some multiple derivative of $Q$ will have both its constant and linear coefficients zero, which means that it has a double root at $0$ , impossible.

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#42 h Mar 22, 2025, 12:03 PM

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...........

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#43 h Mar 23, 2025, 4:51 PM

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KevinYang2.71 wrote:

scannose wrote:

tfw you wrote up a fakesolve that turned out to be correct

KevinYang2.71 wrote:

Claim. $A(x)$ cannot have $n-2$ distinct real roots.

how many points for claiming this but not proving it

2 for everything before using derivative maybe

if only prove the claim，including Rolle，but havent solved others，how many point can i get，thanks

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#44 h Mar 23, 2025, 8:18 PM

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this problem has so much aura

argue by contradiction.
consider $k+1$ real roots $a_1, a_2, .., a_k$ and the $k+1$ factors of degree $k$ . Clearly two of these factors have a 0 in the same place. This means that the gcd of these two factors have two consecutive zeroes.

We can prove by induction that no polynomial with all real roots has two consecutive zeroes. If $Q$ has all real roots with single multiplicity, then $Q'$ does too, as each root of $Q'$ lies in between two roots of $Q$ . But if $Q$ has two consecutive zeroes, so does $Q'$ , contradiction.

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#45 h Mar 24, 2025, 12:43 AM • 1 Y

Y by Ilikeminecraft

Trivial by Advanced Academic Course Precalculus (PNI Chart)

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#46 h Mar 24, 2025, 6:02 AM

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golue3120 wrote:

Here, we have a solution without WLOG $k=n-1$ , without Descartes rule of signs, without Rolle, but is somehow not terrible?

Let $e_k$ denote the degree $k$ elementary symmetric polynomial.

Suppose for the sake of contradiction that all roots of $P(x)$ were real. Let the roots be $r_1,r_2,\dotsc,r_n$ . Let $R$ be the set of roots. Then for every subset $S$ of $k$ roots, $e_i(S)=0$ for some $1\le i\le k-1$ .

Define two subsets of $k$ elements to be adjacent if they share $k-1$ elements.
Lemma. If $\mathcal F$ is a family of $k$ -element subsets of $R$ such that no two are adjacent, then $|\mathcal F|\le\frac{1}{k+1}\binom nk$ .
Proof. Every set $T$ in $\mathcal F$ is adjacent to exactly $k(n-k)$ sets outside of $\mathcal F$ because we may pick any element of $T$ to remove and any root not in $T$ to add. Meanwhile, any set $U$ not in $\mathcal F$ is adjacent to at most $n-k$ sets in $\mathcal F$ because when we remove an element of $U$ and add a root not in $U$ , the additional roots for the adjacent sets in $\mathcal F$ must be distinct. Counting then gives the desired result.

Now define $\mathcal F_i$ to be the family of $k$ -element subsets of roots such that $e_i$ vanishes. Then there are $k-1$ such families, and every size $k$ subset of the roots is in at least one family, so one family must have size at least $\frac{1}{k-1}\binom nk$ . Therefore, at least one family must contain two adjacent subsets. This implies that there are roots $r_1,r_2,\dotsc,r_k,r_k'$ such that
$e_i(r_1,r_2,\dotsc,r_k)=e_i(r_1,r_2,\dotsc,r_{k-1},r_k')=0.$ Now we expand these by writing
$e_i(r_1,r_2,\dotsc,r_{k-1})+r_ke_{i-1}(r_1,r_2,\dotsc,r_{k-1})=e_i(r_1,r_2,\dotsc,r_{k-1})+r_k'e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0,$ and since $r_k\neq r_k'$ , we have
$e_i(r_1,r_2,\dotsc,r_{k-1})=e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0.$
Suppose that $\{r_1,r_2,\dotsc,r_{k-1}\}$ is a set of real numbers such that $e_i(r_1,r_2,\dotsc,r_{k-1})=e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0$ for some $i$ , but no strict subset has this property. It suffices to show that such a set cannot exist.

Now we perform a bit of further expansion
$e_i(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-1}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=0,$ from which we may deduce that
$e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2.$
By minimality, none of $e_i$ , $e_{i-2}$ , or $e_{i-1}$ can vanish, for then two consecutive elementary symmetric polynomials must vanish. By "On the nonnegativity of generalized discriminants of quadratics",
$e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2\ge \frac{i(k-i)}{(i-1)(k-i-1)}e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2}).$ Since $e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2$ is strictly positive, we have a contradiction.

Wow. I spent around two hours in contest looking for an inequality-based approach to proving this but failed. Honestly, given the premise of the problem, I'm slightly surprised one exists. Is there any clean way to prove the inequality you cite? It looks like the terms could line up with some Cauchy-Schwartz / Sum of Squares approach, but I don't immediately see how to do it.
(I ultimately found the Descartes' Rule of Signs approach, so fortunately I avoided a crisis.

)

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#47 h Mar 26, 2025, 4:10 PM • 1 Y

Y by KevinYang2.71

We show the converse that, if all polynomials dividing $P$ have some zero coefficient, $P(x)$ has a nonzero root. Clearly, it suffices to show $k = n - 1$ . WLOG let $a_k = 1$ and WLOG only consider monic degree- $k$ polynomials dividing $P$ . Let, $P(x) = (x - r_1) \dots (x - r_n)$ . Its divisors with degree $n - 1$ have,
$\frac{P(x)}{x - r_1}, \frac{P(x)}{x - r_2}, \dots, \frac{P(x)}{x - r_n}$
FTSOC, let each of these divisors have a zero coefficient. By Pigeonhole, since they are monic degree $n-1$ polynomials, two of them share a zero coefficient in the same spot. WLOG, let $\frac{P(x)}{x - r_1}, \frac{P(x)}{x - r_2}$ both have a coefficient of zero for $x^{n-1-e}$ .

Let $\sigma (S, k)$ be the $k$ -th symmetric sum of a set $S$ . Let $R = \{r_1, \dots, r_n\}$ . By Vieta's,
$\sigma (R - r_1, e) = \sigma (R - r_2, e) = 0$ $\iff r_2 \sigma (R - r_1 - r_2, e - 1) + \sigma (R - r_1 - r_2, e) = r_1 \sigma (R - r_1 - r_2, e - 1) + \sigma (R - r_1 - r_2, e) = 0$
As a result, $r_2 \sigma (R - r_1 - r_2, e - 1) = r_1 \sigma (R - r_1 - r_2, e - 1)$ . Since roots are distinct, we must have $\sigma (R - r_1 - r_2, e - 1) = 0$ . Plugging this back in, $\sigma (R - r_1 - r_2, e) = 0$ as well.

(Above is the extent to which I wrote up my solution on the test)

By Vieta's, $Q(x) = \frac{P(x)}{(x - r_1)(x - r_2)}$ has two consecutive zero coefficients for $x^m, x^{m-1}$ for some $m$ . However, it also has distinct real roots. We show that this is impossible on real polynomials.

(In red is what I couldn't figure out during the test :wallbash_red:

:wallbash_red:

)
Whenever we take the derivative of a polynomial with distinct real roots, which we label $r_1 < r_2 < \dots < r_p$ , by the Mean value theorem, there exists a point on each of $(r_1, r_2), (r_2, r_3), \dots, (r_{p-1}, r_p)$ where the derivative is zero meaning there are $p-1$ instances where the derivative is zero. Therefore, since there cannot be additional roots by FToA, the derivative of the polynomial also has distinct real roots.

Therefore, by induction, if we take the $m-1$ th derivative of $Q(x)$ , we still have a polynomial with distinct roots. However, the coefficient of the $x$ term and the constant term is zero so the polynomial has a double root at zero. (!)

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#48 h Apr 12, 2025, 12:07 PM

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Math4Life2020 wrote:

Wow. I spent around two hours in contest looking for an inequality-based approach to proving this but failed. Honestly, given the premise of the problem, I'm slightly surprised one exists. Is there any clean way to prove the inequality you cite? It looks like the terms could line up with some Cauchy-Schwartz / Sum of Squares approach, but I don't immediately see how to do it.

Well, the inequality we need is just the famous Newton inequality
$S_k^2>S_{k-1}S_{k+1}$ for non-zero $x_1,\dots,x_n$ , which are not all equal. Here, $S_k=\frac{\sigma_k}{\binom{n}{k}}$ is the normalized version of the elementary symmetric polynomials of $x_1,\dots,x_n$ . This of course finishes off the problem immediately, so I am slightly surprised that no one has mentioned it before... :maybe:

:maybe:

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