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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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0 replies
jlacosta
Apr 2, 2025
0 replies
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Factor sums of integers
Aopamy   2
N 6 minutes 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
+1 w
Aopamy
Feb 23, 2023
cadaeibf
6 minutes ago
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Least integer T_m such that m divides gauss sum
Al3jandro0000   33
N 10 minutes 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
10 minutes ago
J
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Estonian Math Competitions 2005/2006
STARS   2
N 12 minutes 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
12 minutes ago
J
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Sum of whose elements is divisible by p
nntrkien   43
N 24 minutes 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
+2 w
nntrkien
Aug 8, 2004
lpieleanu
24 minutes ago
J
No more topics!
H
J
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Unsolved NT, 3rd time posting
GreekIdiot   11
N Apr 4, 2025 by GreekIdiot
Source: own
Solve $5^x-2^y=z^3$ where $x,y,z \in \mathbb Z$
Hint
11 replies
GreekIdiot
Mar 26, 2025
GreekIdiot
Apr 4, 2025
J
Unsolved NT, 3rd time posting
G H J
Reveal topic tags
number theory
modular arithmetic
Exponents
unsolved
L
G H BBookmark kLocked kLocked NReply
Source: own
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GreekIdiot
174 posts
GreekIdiot
#1 h  Mar 26, 2025, 11:40 AM • 1 Y
Y by AlexCenteno2007
Solve $5^x-2^y=z^3$ where $x,y,z \in \mathbb Z$
Hint
This post has been edited 2 times. Last edited by GreekIdiot, Mar 26, 2025, 11:41 AM
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GreekIdiot
174 posts
GreekIdiot
#2 h  Mar 28, 2025, 8:50 AM
Y by
guys bump
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whwlqkd
97 posts
whwlqkd
#3 h  Mar 29, 2025, 1:50 AM
Y by
Bump
It is so hard
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DarkDementor
45 posts
DarkDementor
#4 h  Mar 29, 2025, 2:19 AM
Y by
Wrong mbmbmbmb
This post has been edited 1 time. Last edited by DarkDementor, Apr 4, 2025, 11:54 PM
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whwlqkd
97 posts
whwlqkd
#5 h  Mar 29, 2025, 3:36 AM
Y by
z^3 can be 3(mod 4)
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GreekIdiot
174 posts
GreekIdiot
#6 h  Apr 1, 2025, 5:48 PM
Y by
whwlqkd wrote:
Bump
It is so hard

Yeah lol. I took a known BMO SL problem and tried changing some things. The problem was $5^x-3^y=z^2$ or something similar. I changed the $3$ so that modulo $4$ would not solve it immediately and then the power of $z$ to make it a lot harder. Also just noticed that $8$ has a nice property $d=1(mod \: 2) \implies d^3=d(mod \: 8)$.
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GreekIdiot
174 posts
GreekIdiot
#8 h  Apr 3, 2025, 8:37 PM
Y by
Bumping...
DarkDementor wrote:
Mod 4 then LTE should work.
Mind explaining??
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ektorasmiliotis
102 posts
ektorasmiliotis
#9 h  Apr 3, 2025, 8:43 PM
Y by
I don't think mod 4 and lte works... maybe we should try something different, simple mods might not help enough
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ektorasmiliotis
102 posts
ektorasmiliotis
#10 h  Apr 3, 2025, 8:48 PM
Y by
if mod 4 works with a perfect cube...i'll be surprised
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iniffur
536 posts
iniffur
#11 h  Apr 4, 2025, 12:12 AM
Y by
Setting $5^k=X, ~2^t=Y$ wherein $~~k,~t =0.1,2,3....$

It would appear that the equation with x=odd, y= odd of formula:


$5X^2-2Y^2=Z^3$

comprises a non trivial solution $~(X,Y,Z)= (1, 4, -3)\rightarrow (x,y,z)=(1,5,-3)$

@ below: amendments have been made to meet your objection, Thx. I found the solution merely by inspection. What i am suggesting is merely a possible approach other than modulos (but which could turn out to be tougher than the original).
This post has been edited 2 times. Last edited by iniffur, Apr 4, 2025, 7:03 AM
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mathprodigy2011
317 posts
mathprodigy2011
#12 h  Apr 4, 2025, 2:05 AM
Y by
iniffur wrote:
Setting $~5^x=X,~~2^y=Y,~~$the given problem would boil down to solving:

$X^2-Y^2=Z^3$

$5X^2-Y^2=Z^3$

$X^2-2Y^2=Z^3$

$5X^2-2Y^2=Z^3$

A non trivial solution can be found in the last one $~(X,Y,Z)= (1, 4, -3)\rightarrow (x,y,z)=(1,5,-3)$

i mean you definitely got a solution but you wrote X^2 and Y^2 at the start of your proof but isn't is supposed to be X-Y cuz theres no squares in the original? maybe im a bit confused
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GreekIdiot
174 posts
GreekIdiot
#13 h  Apr 4, 2025, 3:19 PM
Y by
Cool idea... Other solutions are pretty easy to come up with. I wonder how we ll prove uniqueness though.
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