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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Shortest number theory you might've seen in your life
AlperenINAN   13
N 3 minutes ago by lksb
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
13 replies
AlperenINAN
May 11, 2025
lksb
3 minutes ago
n-term Sequence
MithsApprentice   15
N 34 minutes ago by Ilikeminecraft
Source: USAMO 1996, Problem 4
An $n$-term sequence $(x_1, x_2, \ldots, x_n)$ in which each term is either 0 or 1 is called a binary sequence of length $n$. Let $a_n$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to 0, 1, 0 in that order. Let $b_n$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that $b_{n+1} = 2a_n$ for all positive integers $n$.
15 replies
MithsApprentice
Oct 22, 2005
Ilikeminecraft
34 minutes ago
Drawing Triangles Against Your Clone
pieater314159   19
N an hour ago by Ilikeminecraft
Source: 2019 ELMO Shortlist C1
Elmo and Elmo's clone are playing a game. Initially, $n\geq 3$ points are given on a circle. On a player's turn, that player must draw a triangle using three unused points as vertices, without creating any crossing edges. The first player who cannot move loses. If Elmo's clone goes first and players alternate turns, who wins? (Your answer may be in terms of $n$.)

Proposed by Milan Haiman
19 replies
pieater314159
Jun 27, 2019
Ilikeminecraft
an hour ago
Odd digit multiplication
JuanDelPan   12
N an hour ago by Ilikeminecraft
Source: Pan-American Girls' Mathematical Olympiad 2021, P4
Lucía multiplies some positive one-digit numbers (not necessarily distinct) and obtains a number $n$ greater than 10. Then, she multiplies all the digits of $n$ and obtains an odd number. Find all possible values of the units digit of $n$.

$\textit{Proposed by Pablo Serrano, Ecuador}$
12 replies
JuanDelPan
Oct 6, 2021
Ilikeminecraft
an hour ago
No more topics!
Collect ...
luutrongphuc   4
N Apr 26, 2025 by Rayanelba
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ such that:
$$f\left(f(xy)+1\right)=xf\left(x+f(y)\right)$$
4 replies
luutrongphuc
Apr 21, 2025
Rayanelba
Apr 26, 2025
Collect ...
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luutrongphuc
58 posts
#1
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Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ such that:
$$f\left(f(xy)+1\right)=xf\left(x+f(y)\right)$$
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GreekIdiot
276 posts
#2
Y by
Try $P(x,1)$ and isolating $x$ on the $RHS$.
$\boxed{f(x)=\dfrac{1}{x}}$ works.
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megarnie
5611 posts
#3 • 2 Y
Y by Alex-131, KevinYang2.71
The answer is $f(x) = \frac ax$ for any positive real $a$, which clearly works. Now we show it's the only solution. Let $P(x,y)$ be the given assertion.

Claim 1: $f$ is surjective.
Proof: Fix $c$ in the image of $f$ and let $f(a) = c$ for some $a$. Setting $y = \frac ax$ gives that $\frac{f(f(xy) + 1)}{x} = \frac{f(c+1)}{x}$ is in the image of $f$ for all positive reals $x$, which takes on all positive reals. $\square$

$P(x,1): f(f(x) + 1) = xf(x + f(1))$.

So by comparing $P(xy,1)$ and $P(x,y)$, we have\[xy f(xy + f(1)) = x f(x + f(y)) \implies y f(xy + f(1)) = f(x + f(y))\]Let $Q(x,y)$ be this assertion.

For $y \ne 1$, $\frac{f(y) - f(1)}{y - 1}$ is not positive (otherwise we could set $x$ equal to it and get $y = 1$ from $Q(x,y)$), so if $f(y) > f(1)$, then $y < 1$, and if $f(y) < f(1)$, then $y > 1$. So if $y > 1$, then $f(y) \le f(1)$ and if $y < 1$, $f(y) \ge f(1)$.

Thus, we have $xf(x + f(y)) = f(f(xy) + 1) \le f(1)$.

Since $f$ is surjective, means that $f(x+y) \le \frac{f(1)}{x}$ for all positive reals $x,y$.

Claim 2: $f(x) \le \frac{f(1)}{x}$ for all $x$.
Proof: Suppose $f(c) > \frac{f(1)}{c}$ for some $c$. Setting $y = c - x$ gives that $f(c) \le \frac{f(1)}{x}$ for $x < c$. Setting $x$ arbitrarily close to $c$ gives a contradiction. $\square$

Claim 3: $f(f(x)) \ge x$ for all $x$.
Proof: Suppose for some $c$ we had $f(f(c)) < c$.

$Q(f(c), c - f(f(c))): f(c) f((c - f(f(c)) ) f(c) + f(1)) = f(c)$, so\[ f((c - f(f(c)) ) f(c) + f(1))  = 1\]However, for any $k > f(1)$, we have $f(k) \le \frac{f(1)}{k} < 1$, a contradiction by setting $k = (c - f(f(c)) ) f(c) + f(1)$. $\square$

Now, we have by setting $x = f(1)$ in claim 3 that $f(f(1)) \le 1$, so combined with $f(f(1)) \ge 1$, we have $f(f(1)) = 1$. Now write $a = f(1)$. We have\[ y f(xy + a) = f(x + f(y)),\]$f(a) = 1$, and also $f(x) \le \frac ax$ for all $x \in \mathbb R^{+}$.

Claim: $f(y) < a$ for all positive reals $y > 1$
Proof: We already know that $f(y) \le a$. If $f(y) = a$, then $f(f(y)) = 1 < y$, absurd by claim 3. $\square$

For $y > 1$, $Q(a - f(y), y): y f((a - f(y)) y + a) = 1$, so\[ f((a - f(y)) y + a) = \frac 1y\]This means\[\frac 1y \le \frac{a}{(a - f(y)) y + a} \implies y \ge y \left( 1 - \frac{f(y)}{a} \right)  + 1\]Therefore, $y \cdot \frac{f(y)}{a} \ge 1$, so $yf(y) \ge a \implies f(y) \ge \frac ay$, but since we already have $f(y)\le \frac ay$, we have $f(y) = \frac ay$ for each $y > 1$, so $f(y) = \frac ay \forall y \ge 1$.

Now, in $Q(x,y)$, fix $y$ to be any positive real number and $x$ large enough so that $xy + a > 1$.

We have $\frac{ay}{xy + a} = \frac{a}{x + f(y)}$, so\[ \frac{ay}{xy + a} = \frac{ay}{(x+f(y)) y} \implies xy + a = xy + y f(y),\]so $y f(y) = a \implies f(y) = \frac ay$ for all positive reals $y$.
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KevinYang2.71
428 posts
#4 • 3 Y
Y by megarnie, Alex-131, FPilipovic
The answer is $f(x)=\frac{a}{x}$ for a constant $a\in\mathbb{R}^+$, which works.

Let $P(x,\,y)$ denote the assertion. From $P(x,\,1/x)$ we get $f(f(1)+1)=xf(x+f(1/x))$ so $f(x+f(1/x))=\frac{f(f(1)+1)}{x}$. Thus $f$ is surjective.

Claim. $f$ is non-increasing.
Proof. Comparing $P(xy,\,z)$ and $P(xz,\,y)$ yields $xyf(xy+f(z))=f(f(xyz)+1)=xzf(xz+f(y))$ so $yf(xy+f(z))=zf(xz+f(y))$. If $\frac{f(y)-f(z)}{y-z}>0$, then $x:=\frac{f(y)-f(z)}{y-z}$ satisfies $xy+f(z)=xz+f(y)$ so $y=z$, a contradiction. Thus $f(y)\leq f(z)$ for all $y>z>0$. $\square$

For each $c\in\mathbb{R}^+$, if $\lim_{x\to c^+}f(x)=b<f(c)$, then all $x>c$ satisfies $f(x)\leq b$ and all $x<c$ satisfies $f(x)\geq f(c)$ so $(b,\,f(c))$ is not in the range of $f$, contradicting surjectivity. Hence $\lim_{x\to c^+}f(x)=f(c)$. Clearly we must have $\lim_{x\to\infty}f(x)=0$, as otherwise a positive number arbitrarily close to $0$ would not be in the range of $f$. Thus $\lim_{y\to\infty}f(f(xy)+1)=f(1)$ and $\lim_{y\to\infty}xf(x+f(y))=xf(x)$ so $f(x)=\frac{f(1)}{x}$, as desired. $\square$
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Rayanelba
20 posts
#5
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Veryyy cool fe
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