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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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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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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
Easy function in turkey TST
egxa   10
N 9 minutes ago by jasperE3
Source: 2024 Turkey TST P2
Find all $f:\mathbb{R}\to\mathbb{R}$ functions such that
$$f(x+y)^3=(x+2y)f(x^2)+f(f(y))(x^2+3xy+y^2)$$for all real numbers $x,y$
10 replies
egxa
Mar 18, 2024
jasperE3
9 minutes ago
problem//
Cobedangiu   3
N 10 minutes ago by Cobedangiu
Let $x,y,z>0$ and $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3$. Find min A (and prove)
$A=\sum \dfrac{1}{\sqrt{2x^2+y^2+3}}$
3 replies
Cobedangiu
2 hours ago
Cobedangiu
10 minutes ago
Turbo's en route to visit each cell of the board
Lukaluce   9
N 17 minutes ago by CrazyInMath
Source: EGMO 2025 P5
Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^2$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate $90^{\circ}$ counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of $n$, the maximum number of good cells over all possible starting configurations.

Proposed by Melek Güngör, Turkey
9 replies
Lukaluce
Today at 11:01 AM
CrazyInMath
17 minutes ago
EGMO magic square
Lukaluce   4
N 20 minutes ago by Polyquadratus
Source: EGMO 2025 P6
In each cell of a $2025 \times 2025$ board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to $1$, and the sum of the numbers in each column is equal to $1$. Define $r_i$ to be the largest value in row $i$, and let $R = r_1 + r_2 + ... + r_{2025}$. Similarly, define $c_i$ to be the largest value in column $i$, and let $C = c_1 + c_2 + ... + c_{2025}$.
What is the largest possible value of $\frac{R}{C}$?

Proposed by Paulius Aleknavičius, Lithuania
4 replies
Lukaluce
6 hours ago
Polyquadratus
20 minutes ago
No more topics!
NMO (Nepal) Problem 4
khan.academy   8
N Apr 4, 2025 by godchunguus
Find all integer/s $n$ such that $\displaystyle{\frac{5^n-1}{3}}$ is a prime or a perfect square of an integer.

Proposed by Prajit Adhikari, Nepal
8 replies
khan.academy
Mar 17, 2024
godchunguus
Apr 4, 2025
NMO (Nepal) Problem 4
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khan.academy
634 posts
#1 • 1 Y
Y by ItsBesi
Find all integer/s $n$ such that $\displaystyle{\frac{5^n-1}{3}}$ is a prime or a perfect square of an integer.

Proposed by Prajit Adhikari, Nepal
This post has been edited 1 time. Last edited by khan.academy, Mar 17, 2024, 2:39 PM
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vsamc
3788 posts
#2 • 3 Y
Y by khan.academy, vsarg, lamphead
Solution
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Basu_Dev
17 posts
#3 • 1 Y
Y by khan.academy
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This post has been edited 1 time. Last edited by Basu_Dev, Mar 17, 2024, 2:43 PM
Reason: Code error
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Taha-R
45 posts
#4 • 1 Y
Y by khan.academy
Answer : $ n = 0 $.
Solution :

Case 1 : $ \frac {5^n - 1} {3} = p $ (prime).
We know that $\frac {5^n - 1} {3} = \frac {4.(5^{n-1} + ... + 1)} {3} = p.$ But we know that $ gcd(3,4) = 1 \Rightarrow 4 | p $. Contradiction.
Case 2 : $ \frac {5^n - 1} {3} = a^2 \Rightarrow 5^n -1 = 3a^2 $.
We khow that $ a^2 = {-1 , 0 ,1}  $(mod 5) . But $ LHS = 4  $ (mod 5) $ \Rightarrow $ Contradiction unless $ n = 0 $.
This post has been edited 4 times. Last edited by Taha-R, Mar 17, 2024, 4:48 PM
Reason: .
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Jishnu4414l
154 posts
#5
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We claim that $n=0$ is the only solution.
First let $\frac{5^{n-1}}{3}=q^2$
In this case we have $5^n-1=3q^2$. Viewing the equation modulo $5$, we see that there are no possible solutions.
Now let $\frac{5^{n-1}}{3}=p$ ($p$ is a prime)
As $5^{n-1} \equiv 0 \pmod 4$ (for $n\geq 1$), we see that $\frac{5^{n-1}}{3}$ if an integer is always $0  \pmod 4$ \for $n\geq 1$.
We see that $n=0$ works and thus it is the only possible solution.
This post has been edited 1 time. Last edited by Jishnu4414l, Mar 28, 2024, 1:12 PM
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AshAuktober
985 posts
#6
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Sketch: if prime then has to be 2, impossible; now n>= 0 ofc and n = 0 work so consider n >0 then mod 5 kills
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pie854
243 posts
#7
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This is $0\pmod 4$ and $3\pmod 5$ and hence it cannot be a prime or a perfect square.
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cubres
111 posts
#8 • 1 Y
Y by XAN4
Solution
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godchunguus
13 posts
#9
Y by
First, try $n = 0$, which clearly works. Then, we claim that no such $n\ge1$ exists.

$\boxed{\textbf{Let }p\textbf{ denote any prime, then, }}$

$\frac{5^n-1}{3}=p$
$5^n-1 \ = \ 3p$
$0 \equiv 3p$ (mod 4)
$ p \equiv 0$ (mod 4)
Which is clearly ridiculous. Hence, no such n exists for $\frac{5^n-1}{3}$ to be a prime.

$\boxed{\textbf{Let }k\textbf{ denote any perfect square, then, }}$

$\frac{5^n-1}{3} = k$
$5^n-1 = 3k$
$-1 \equiv 3k$ (mod 5)

Which, again, is ridiculous. Hence, our claim follows, leading us to no solutions for $n$, except $n=0$.
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