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pomodor_ap   4
N 37 minutes ago by jasperE3
Source: Own, PDC001-P7
Let $f : \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that
$$f(x)f(x^2 + y f(y)) = f(x)f(y^2) + x^3$$for all $x, y \in \mathbb{R}^+$. Determine all such functions $f$.
4 replies
pomodor_ap
Yesterday at 11:24 AM
jasperE3
37 minutes ago
J
H
FE solution too simple?
Yiyj1   8
N 44 minutes ago by lksb
Source: 101 Algebra Problems from the AMSP
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
8 replies
Yiyj1
Apr 9, 2025
lksb
44 minutes ago
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Polynomials in Z[x]
BartSimpsons   16
N an hour ago by bin_sherlo
Source: European Mathematical Cup 2017 Problem 4
Find all polynomials $P$ with integer coefficients such that $P (0)\ne 0$ and $$P^n(m)\cdot P^m(n)$$is a square of an integer for all nonnegative integers $n, m$.

Remark: For a nonnegative integer $k$ and an integer $n$, $P^k(n)$ is defined as follows: $P^k(n) = n$ if $k = 0$ and $P^k(n)=P(P(^{k-1}(n))$ if $k >0$.

Proposed by Adrian Beker.
16 replies
BartSimpsons
Dec 27, 2017
bin_sherlo
an hour ago
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Why is the old one deleted?
EeEeRUT   13
N an hour ago by EVKV
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
13 replies
EeEeRUT
Apr 16, 2025
EVKV
an hour ago
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Factor sums of integers
Aopamy   2
N 2 hours ago by cadaeibf
Let $n$ be a positive integer. A positive integer $k$ is called a benefactor of $n$ if the positive divisors of $k$ can be partitioned into two sets $A$ and $B$ such that $n$ is equal to the sum of elements in $A$ minus the sum of the elements in $B$. Note that $A$ or $B$ could be empty, and that the sum of the elements of the empty set is $0$.

For example, $15$ is a benefactor of $18$ because $1+5+15-3=18$.

Show that every positive integer $n$ has at least $2023$ benefactors.
2 replies
Aopamy
Feb 23, 2023
cadaeibf
2 hours ago
J
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Least integer T_m such that m divides gauss sum
Al3jandro0000   33
N 2 hours ago by NerdyNashville
Source: 2020 Iberoamerican P2
Let $T_n$ denotes the least natural such that
$$n\mid 1+2+3+\cdots +T_n=\sum_{i=1}^{T_n} i$$Find all naturals $m$ such that $m\ge T_m$.

Proposed by Nicolás De la Hoz
33 replies
Al3jandro0000
Nov 17, 2020
NerdyNashville
2 hours ago
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Estonian Math Competitions 2005/2006
STARS   2
N 2 hours ago by jasperE3
Source: Juniors Problem 4
A $ 9 \times 9$ square is divided into unit squares. Is it possible to fill each unit square with a number $ 1, 2,..., 9$ in such a way that, whenever one places the tile so that it fully covers nine unit squares, the tile will cover nine different numbers?
2 replies
STARS
Jul 30, 2008
jasperE3
2 hours ago
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Sum of whose elements is divisible by p
nntrkien   43
N 2 hours ago by lpieleanu
Source: IMO 1995, Problem 6, Day 2, IMO Shortlist 1995, N6
Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?
43 replies
nntrkien
Aug 8, 2004
lpieleanu
2 hours ago
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Arrangement of integers in a row with gcd
egxa   2
N 2 hours ago by Qing-Cloud
Source: All Russian 2025 10.5 and 11.5
Let \( n \) be a natural number. The numbers \( 1, 2, \ldots, n \) are written in a row in some order. For each pair of adjacent numbers, their greatest common divisor (GCD) is calculated and written on a sheet. What is the maximum possible number of distinct values among the \( n - 1 \) GCDs obtained?
2 replies
egxa
Apr 18, 2025
Qing-Cloud
2 hours ago
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Integer representation
RL_parkgong_0106   1
N 2 hours ago by Jackson0423
Source: Own
Show that for any positive integer $n$, there exists some positive integer $k$ that makes the following equation have no integer root $(x_1, x_2, x_3, \dots, x_n)$.

$$x_1^{2^1}+x_2^{2^2}+x_3^{2^3}+\dots+x_n^{2^n}=k$$
1 reply
RL_parkgong_0106
4 hours ago
Jackson0423
2 hours ago
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Geometry :3c
popop614   4
N Apr 4, 2025 by goaoat
Source: MINE :<
Quadrilateral $ABCD$ has an incenter $I$ Suppose $AB > BC$. Let $M$ be the midpoint of $AC$. Suppose that $MI \perp BI$. $DI$ meets $(BDM)$ again at point $T$. Let points $P$ and $Q$ be such that $T$ is the midpoint of $MP$ and $I$ is the midpoint of $MQ$. Point $S$ lies on the plane such that $AMSQ$ is a parallelogram, and suppose the angle bisectors of $MCQ$ and $MSQ$ concur on $IM$.

The angle bisectors of $\angle PAQ$ and $\angle PCQ$ meet $PQ$ at $X$ and $Y$. Prove that $PX = QY$.
4 replies
popop614
Apr 3, 2025
goaoat
Apr 4, 2025
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Geometry :3c
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Source: MINE :<
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popop614
271 posts
popop614
#1 h  Apr 3, 2025, 12:19 AM • 3 Y
Y by sixoneeight, OronSH, goaoat
Quadrilateral $ABCD$ has an incenter $I$ Suppose $AB > BC$. Let $M$ be the midpoint of $AC$. Suppose that $MI \perp BI$. $DI$ meets $(BDM)$ again at point $T$. Let points $P$ and $Q$ be such that $T$ is the midpoint of $MP$ and $I$ is the midpoint of $MQ$. Point $S$ lies on the plane such that $AMSQ$ is a parallelogram, and suppose the angle bisectors of $MCQ$ and $MSQ$ concur on $IM$.

The angle bisectors of $\angle PAQ$ and $\angle PCQ$ meet $PQ$ at $X$ and $Y$. Prove that $PX = QY$.
This post has been edited 1 time. Last edited by popop614, Apr 3, 2025, 12:42 AM
Reason: asfdasdf
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sixoneeight
1138 posts
sixoneeight
#2 h  Apr 3, 2025, 12:42 AM
Y by
popop614 wrote:
$MI \perp BI$. $DI$
mibidi
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Ianis
402 posts
Ianis
#3 h  Apr 3, 2025, 12:45 AM
Y by
sixoneeight wrote:
popop614 wrote:
$MI \perp BI$. $DI$
mibidi

skibidi?
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ItzsleepyXD
108 posts
ItzsleepyXD
#4 h  Apr 3, 2025, 3:28 AM
Y by
Ianis wrote:
sixoneeight wrote:
popop614 wrote:
$MI \perp BI$. $DI$
mibidi

skibidi?

dop dop yes yes
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goaoat
94 posts
goaoat
#5 h  Apr 4, 2025, 8:00 PM • 2 Y
Y by mpcnotnpc, sixoneeight
sogood :love:

$\textbf{Dealing with S:}$
First, let's eliminate $S$. Note that the angle bisector concurrence is the same as $$AQ/AM=MS/SQ=MC/CQ=AM/CQ \implies AM^2=AQ \cdot CQ$$Furthermore, the $PX=QY$ relation is the same as $AQ/AP=CP/CQ$ by angle bisector theorem, which we can simplify to $AM^2=AP \cdot CP$ using the previous equation. Thus, it suffices to prove this last relation.

$\textbf{Dealing with Q:}$
The equality $AM^2=AM \cdot AC=AQ \cdot CQ$ implies $Q$ lies on the lemniscate of Bernoulli with foci $A$ and $C$, $\mathcal{L}$. Thus, the inverse of $Q$ with respect to $(AC)$ lies on the rectangular hyperbola with foci $A$ and $C$, $\mathcal{H}$.

Now, the pedal curve of $\mathcal{H}$ is the image of $\mathcal{L}$ after a $\times 1/2$ homothety at $M$ . This gives that the perpendicular bisector of $MQ$, or $BI$, is tangent to $\mathcal{H}$. However, since we have $\angle ABI = \angle CBI$, by the optical property, the unique point on $BI$ that satisfies this is its tangency with $\mathcal{H}$. Consequently, $B$ lies on $\mathcal{H}$.

Now, $B$ lying on $\mathcal{H}$ is equivalent to $|AB-BC|=\frac{2+\sqrt{2}}{4} AC$, and this can be rearranged with the median length formula to $AB \cdot BC = BM^2$.

Let $Q'$ be the harmonic conjugate of $B$ with respect to $(BAC)$. Then, $$BC/BQ'=BM/AB \implies AB \cdot BC = BM \cdot BQ'=BM^2$$Thus, $BQ'=BM$. Because $Q'$ also lies on the $B$-symmedian, we conclude $Q=Q'$ so $Q$ lies on $(BAC)$.

$\textbf{Dealing with D:}$
Furthermore, $ABCD$ is tangential so $AB+CD=AD+BC$ or $AB-BC=AD-CD$, and this implies $D$ lies on $\mathcal{H}$.

Claim: $\angle MBD = 90^\circ$

Proof: Let the reflection of $B$ over $M$ be $B'$. Note $B'$ lies on $\mathcal{H}$ as $M$ is its center. Let $N$ be the midpoint of $BD$. Realize $MN$ is the Newton line of $ABCD$ so $I$ lies on it. So, $BI \perp (B'D \parallel MN)$. Since triangle $B'BD$ lies on $\mathcal{H}$, its altitude $BI$ should intersect $\mathcal{H}$ at its orthocenter, but $BI$ is tangent to $\mathcal{H}$. Ergo, $B$ is the orthocenter of triangle $B'BD$, proving the claim.

$\textbf{Dealing with P:}$
The previous claim gives $\angle MTD = 90^\circ$, so $P$ is the point on the $D$-symmedian of $DAC$ satisfying $DM=DP$.

Using the same phantom point argument we did with $Q$, we can show that $P$ is the harmonic conjugate of $D$ with respect to $(DAC)$.

Claim: The inverse of $P$ with respect to $(AC)$ is $D'$, the reflection of $D$ over $AC$.

Proof: It's easy to show that the inverse of $(DAC)$ with respect to $(AC)$ is $(D'AC)$, the reflection of $(DAC)$ over $M$. Let $D''$ be the reflection of $D$ over the perpendicular bisector of $AC$. Note $D''$ is the intersection of $PD'$ with $(DAC)$ by some projective, so the inverse of $P$ is the reflection of $D''$ over $M$ which is $D'$ (it is not the reflection of $P$ over $M$ as they lie on different sides of $M$).

$\textbf{Finish:}$
Finally, $D'$ lies on $\mathcal{H}$ as $\mathcal{H}$ is symmetric about $AC$, so $P$ lies on $\mathcal{L}$. This means $AP \cdot CP = AM \cdot CM = AM^2$, which is what we needed to prove, so we are done.
This post has been edited 3 times. Last edited by goaoat, Apr 4, 2025, 8:14 PM
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