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Integer polynomial commutes with sum of digits
cjquines0   45
N 41 minutes ago by anudeep
Source: 2016 IMO Shortlist N1
For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2016$, the integer $P(n)$ is positive and $$S(P(n)) = P(S(n)).$$
Proposed by Warut Suksompong, Thailand
45 replies
cjquines0
Jul 19, 2017
anudeep
41 minutes ago
Good Permutations in Modulo n
swynca   11
N an hour ago by Assassino9931
Source: BMO 2025 P1
An integer $n > 1$ is called $\emph{good}$ if there exists a permutation $a_1, a_2, a_3, \dots, a_n$ of the numbers $1, 2, 3, \dots, n$, such that:
$(i)$ $a_i$ and $a_{i+1}$ have different parities for every $1 \leq i \leq n-1$;
$(ii)$ the sum $a_1 + a_2 + \cdots + a_k$ is a quadratic residue modulo $n$ for every $1 \leq k \leq n$.
Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.
11 replies
swynca
Apr 27, 2025
Assassino9931
an hour ago
Israel Number Theory
mathisreaI   67
N an hour ago by lolsamo
Source: IMO 2022 Problem 5
Find all triples $(a,b,p)$ of positive integers with $p$ prime and \[ a^p=b!+p. \]
67 replies
mathisreaI
Jul 13, 2022
lolsamo
an hour ago
IMO 2016 Shortlist, N6
dangerousliri   69
N an hour ago by MR.1
Denote by $\mathbb{N}$ the set of all positive integers. Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for all positive integers $m$ and $n$, the integer $f(m)+f(n)-mn$ is nonzero and divides $mf(m)+nf(n)$.

Proposed by Dorlir Ahmeti, Albania
69 replies
dangerousliri
Jul 19, 2017
MR.1
an hour ago
Number of Solutions is 2
Miku3D   30
N 2 hours ago by lakshya2009
Source: 2021 APMO P1
Prove that for each real number $r>2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^2=r\lfloor x \rfloor$.
30 replies
Miku3D
Jun 9, 2021
lakshya2009
2 hours ago
Problem 3
blug   3
N 2 hours ago by sunken rock
Source: Czech-Polish-Slovak Junior Match 2025 Problem 3
In a triangle $ABC$, $\angle ACB=60^{\circ}$. Points $D, E$ lie on segments $BC, AC$ respectively. Points $K, L$ are such that $ADK$ and $BEL$ are equlateral, $A$ and $L$ lie on opposite sides of $BE$, $B$ and $K$ lie on the opposite siedes of $AD$. Prove that
$$AE+BD=KL.$$
3 replies
blug
Yesterday at 4:47 PM
sunken rock
2 hours ago
Computing functions
BBNoDollar   6
N 2 hours ago by BBNoDollar
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
6 replies
BBNoDollar
May 18, 2025
BBNoDollar
2 hours ago
Hard Inequality
danilorj   5
N 3 hours ago by danilorj
Let $a, b, c > 0$ with $a + b + c = 1$. Prove that:
\[
\sqrt{a + (b - c)^2} + \sqrt{b + (c - a)^2} + \sqrt{c + (a - b)^2} \geq \sqrt{3},
\]with equality if and only if $a = b = c = \frac{1}{3}$.
5 replies
danilorj
Today at 5:17 AM
danilorj
3 hours ago
easy geo
ErTeeEs06   5
N 3 hours ago by Adywastaken
Source: BxMO 2025 P3
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$. Let $D, E, F$ be the midpoints of the arcs $\stackrel{\frown}{BC}, \stackrel{\frown}{CA}, \stackrel{\frown}{AB}$ of $\Omega$ not containing $A, B, C$ respectively. Let $D'$ be the point of $\Omega$ diametrically opposite to $D$. Show that $I, D'$ and the midpoint $M$ of $EF$ lie on a line.
5 replies
ErTeeEs06
Apr 26, 2025
Adywastaken
3 hours ago
Inspired by SXJX (12)2022 Q1167
sqing   4
N 3 hours ago by sqing
Source: Own
Let $ a,b,c>0 $. Prove that$$\frac{kabc-1} {abc(a+b+c+8(2k-1))}\leq \frac{1}{16 }$$Where $ k>\frac{1}{2}.$
4 replies
sqing
Yesterday at 4:01 AM
sqing
3 hours ago
Geometry hard problem.
noneofyou34   3
N 3 hours ago by noneofyou34
In a circle of radius R, three chords of length R are given. Their ends are joined with segments to
obtain a hexagon inscribed in the circle. Show that the midpoints of the new chords are the vertices of
an equilateral triang
3 replies
noneofyou34
Today at 9:50 AM
noneofyou34
3 hours ago
Computational polynomial
kamatadu   5
N Jan 12, 2025 by S_14159
Source: STEMS 2023 Maths CAT A Part A P5
Consider a polynomial $P(x) \in \mathbb{R}[x]$, with degree $2023$, such that $P(\sin^2(x))+P(\cos^2(x)) =1$ for all $x \in \mathbb{R}$. If the sum of all roots of $P$ is equal to $\dfrac{p}{q}$ with $p, q$ coprime, then what is the product $pq$?
5 replies
kamatadu
Jan 8, 2023
S_14159
Jan 12, 2025
Computational polynomial
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G H BBookmark kLocked kLocked NReply
Source: STEMS 2023 Maths CAT A Part A P5
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kamatadu
480 posts
#1 • 3 Y
Y by HoripodoKrishno, Rounak_iitr, Mango247
Consider a polynomial $P(x) \in \mathbb{R}[x]$, with degree $2023$, such that $P(\sin^2(x))+P(\cos^2(x)) =1$ for all $x \in \mathbb{R}$. If the sum of all roots of $P$ is equal to $\dfrac{p}{q}$ with $p, q$ coprime, then what is the product $pq$?
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starchan
1610 posts
#2
Y by
Some work shows that if $t$ is a root of $P(x)-1/2$ then so is $1-t$. Since $\deg P$ is odd, this forces one root of $P(x)-1/2$ to be $1/2$ and the others can be paired up into roots summing to one. Thus sum of roots is $2023/2$ and answer is $4046$.
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pco
23515 posts
#3 • 2 Y
Y by StarLex1, Mango247
kamatadu wrote:
Consider a polynomial $P(x) \in \mathbb{R}[x]$, with degree $2023$, such that $P(\sin^2(x))+P(\cos^2(x)) =1$ for all $x \in \mathbb{R}$. If the sum of all roots of $P$ is equal to $\dfrac{p}{q}$ with $p, q$ coprime, then what is the product $pq$?
Other method :
Since polynomial, this implies $P(x)+P(1-x)=1$ $\forall x$

Let $P(x)=ax^{2023}+bx^{2022}+... $ (with $a\ne 0$)

Coefficient pf $x^{2022}$ in $P(x)+P(1-x)$ is $2023a+2b$ and must be zero and so $-\frac ba=\frac{2023}2$

Hence result $\boxed{4046}$
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chakrabortyahan
385 posts
#4
Y by
This problem resembles with some old USATST
$P(x)+P(1-x)-1 = Q(x) $ now note that $Q$ is a zero poly.
So , $P(x)+P(1-x) \equiv 1 $
now let the roots of $P$ be $\alpha_ i $ so the roots of $P(x)=1 $ are also $(1-\alpha_i)$s .Summing we get $\text{sum of the roots}= \frac{2003}{2}$
This post has been edited 1 time. Last edited by chakrabortyahan, Jan 8, 2023, 3:49 PM
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lifeismathematics
1188 posts
#5
Y by
nice
This post has been edited 1 time. Last edited by lifeismathematics, Jan 11, 2023, 3:51 AM
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S_14159
48 posts
#6
Y by
It is easy to see that $P(x)+P(1-x)=1$, taking $P(x)=\sum_{i=0}^{2023}a_{i}x^{i}$ we see that coefficient of $x^{2022}$ is $2023 a+2b=0$ which basically gives us $\boxed{\frac{-b}{a}=\frac{2023}{2}}$
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