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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Number Theory
AnhQuang_67   3
N a minute ago by GreekIdiot
Source: HSGSO 2024
Let $p$ be an odd prime number and a sequence $\{a_n\}_{n=1}^{+\infty}$ satisfy $$a_1=1, a_2=2$$and $$a_{n+2}=2\cdot a_{n+1}+3\cdot a_n, \forall n \geqslant 1$$Prove that always exists positive integer $k$ satisfying for all positive integers $n$, then $a_n \ne k \mod{p}$.

P/s: $\ne$ is "not congruence"
3 replies
AnhQuang_67
2 hours ago
GreekIdiot
a minute ago
Solve All 6 IMO 2024 Problems (42/42), New Framework Looking for Feedback
Blackhole.LightKing   3
N 9 minutes ago by DottedCaculator
Hi everyone,

I’ve been experimenting with a different way of approaching mathematical problem solving — a framework that emphasizes recursive structures and symbolic alignment rather than conventional step-by-step strategies.

Using this method, I recently attempted all six problems from IMO 2024 and was able to arrive at what I believe are valid full-mark solutions across the board (42/42 total score, by standard grading).

However, I don’t come from a formal competition background, so I’m sure there are gaps in clarity, communication, or even logic that I’m not fully aware of.

If anyone here is willing to take a look and provide feedback, I’d appreciate it — especially regarding:

The correctness and completeness of the proofs

Suggestions on how to make the ideas clearer or more elegant

Whether this approach has any broader potential or known parallels

I'm here to learn more and improve the presentation and thinking behind the work.

You can download the Solution here.

https://agi-origin.com/assets/pdf/AGI-Origin_IMO_2024_Solution.pdf


Thanks in advance,
— BlackholeLight0


3 replies
+1 w
Blackhole.LightKing
3 hours ago
DottedCaculator
9 minutes ago
2 var inequalities
sqing   3
N 12 minutes ago by sqing
Source: Own
Let $ a,b> 0 $ and $ a+b\leq  2ab . $ Prove that
$$ \frac{   a + b }{ a^2(1+ b^2)} \leq \sqrt 5-1$$$$ \frac{   a +ab+ b }{ a^2(1+ b^2)} \leq \frac{3(\sqrt5-1)}{2}$$$$ \frac{   a +a^2b^2+ b }{ a^2(1+ b^2)} \leq2$$Solution:
$a\ge\frac{b}{2b-1}, b>\frac12$ and $  \frac{   a +a^2b^2+ b }{ a^2(1+ b^2)} \le\frac{2ab+a^2b^2}{a^2(1+b^2)}=1+\frac{2b-a}{a(1+b^2)} \le 1+\frac{4b-3}{b^2+1}$

Assume $u=4b-3>0$ then $  \frac{   a +a^2b^2+ b }{ a^2(1+ b^2)}  \le 1+\frac{16u}{u^2+6u+25} =2+ \frac{16}{6+u+\frac{25}u} \le 3$
Equalityholds when $a=\frac{2}{3},b=2. $
3 replies
sqing
Yesterday at 1:13 PM
sqing
12 minutes ago
hard problem
Cobedangiu   8
N 12 minutes ago by ReticulatedPython
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
8 replies
Cobedangiu
Apr 21, 2025
ReticulatedPython
12 minutes ago
No more topics!
Two Functional Inequalities
Mathdreams   7
N Apr 7, 2025 by John_Mgr
Source: 2025 Nepal Mock TST Day 2 Problem 2
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x) \le x^3$$and $$f(x + y) \le f(x) + f(y) + 3xy(x + y)$$for any real numbers $x$ and $y$.

(Miroslav Marinov, Bulgaria)
7 replies
Mathdreams
Apr 6, 2025
John_Mgr
Apr 7, 2025
Two Functional Inequalities
G H J
Source: 2025 Nepal Mock TST Day 2 Problem 2
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Mathdreams
1466 posts
#1 • 4 Y
Y by PikaPika999, AlexCenteno2007, cubres, khan.academy
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x) \le x^3$$and $$f(x + y) \le f(x) + f(y) + 3xy(x + y)$$for any real numbers $x$ and $y$.

(Miroslav Marinov, Bulgaria)
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grupyorum
1413 posts
#2 • 2 Y
Y by PikaPika999, cubres
I will show $f(x)=x^3$ for all $x$, which clearly works.

Taking $x=y=0$ in both, we find $f(0)\le 0$ and $f(0)\ge 0$, so $f(0)=0$. Taking $y=-x$ in the second, we find $0=f(0)\le f(x)+f(-x)$. Likewise, $f(x)\le x^3$ and $f(-x)\le (-x)^3=-x^3$. So, $f(x)+f(-x)\le 0$. Combining, $f(x)+f(-x)=0$ for all $x$. Lastly, using $f(-x)=-f(x)$, we get $-x^3 \ge f(-x)=-f(x)$, so $f(x)\ge x^3$ too. Thus, $f(x)=x^3$ for all $x$.
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kokcio
69 posts
#3 • 2 Y
Y by PikaPika999, cubres
Putting $x=y=0$, we have $f(0)\leq 2f(0)$, but $f(0)\leq 0$, so we have to have that $f(0)=0$.
Putting now $y=-x$, we have $0\leq f(x)+f(-x) \leq x^3 + (-x)^3 = 0$, but this means that we have to have $f(x)=x^3$, because of inequalities.
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John_Mgr
67 posts
#4 • 2 Y
Y by PikaPika999, cubres
Claim: $f(x) \equiv x^3$ is the only function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies the given conditions.
Let P(x,y) be the assertion for the convenience.
i.e $f(x+y) \le f(x)+f(y)+3xy(x+y)$
P(0,0): $f(0)\ge0$ and $f(0)\le0$, $\Rightarrow$ $f(0)=0$,
Assume $f(x)=g(x)+x^3$, $\implies$ $g(x+y)\le g(x)+g(y)$
This tells us that g is subadditive.
$g(x)\le 0$ as $f(x)\le x^3$
Let Q(x,y) be the assertion for $g(x+y)\le g(x)+g(y)$
Q(x,-x): $g(0)\le g(x)+g(-x)$$\rightarrow g(x)+g(-x)\ge 0$ as $f(0)=0$ but $g(x)+g(-x)\le 0$
So, $g(x)+g(-x)=0$$\Rightarrow g(-x)=-g(x) \ge 0$
The only way both $g(x)\le x$ and $g(-x)\ge 0$ hold is if $g(x)=0$, $\forall$ $x\in \mathbb{R}$
Therefore $\boxed{f(x)=x^3}$, $\forall$ $x\in \mathbb{R}$
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jasperE3
11241 posts
#5 • 5 Y
Y by PikaPika999, Assassino9931, megarnie, navier3072, cubres
Let $g(x)=f(x)-x^3+x$, it reduces to this old problem:
https://artofproblemsolving.com/community/c6h257315p1403063
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Maximilian113
554 posts
#6 • 2 Y
Y by PikaPika999, cubres
Note that $x=y=0 \implies 0 \leq f(0) \leq 0 \implies f(0)=0.$ Now $$y=-x \implies 0 \leq f(x)+f(-x) \leq f(x) -x^3 \implies x^3 \leq f(x) \implies f(x)=x^3$$for all $x,$ which clearly works.
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Assassino9931
1251 posts
#7 • 1 Y
Y by cubres
jasperE3 wrote:

Nice catch haha
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John_Mgr
67 posts
#8 • 1 Y
Y by cubres
jasperE3 wrote:

Good one!!
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N Quick Reply
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