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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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0 replies
jlacosta
May 1, 2025
0 replies
J
H
What is thiss
EeEeRUT   3
N a few seconds ago by ItzsleepyXD
Source: Thailand MO 2025 P6
Find all function $f: \mathbb{R}^+ \rightarrow \mathbb{R}$,such that the inequality $$f(x) + f(\frac{y}{x}) \leqslant \frac{x^3}{y^2} + \frac{y}{x^3}$$holds for all positive reals $x,y$ and for every positive real $x$, there exist positive reals $y$, such that the equality holds.
3 replies
+1 w
EeEeRUT
3 hours ago
ItzsleepyXD
a few seconds ago
J
H
Collinearity of intersection points in a triangle
MathMystic33   2
N 13 minutes ago by AylyGayypow009
Source: 2025 Macedonian Team Selection Test P1
On the sides of the triangle \(\triangle ABC\) lie the following points: \(K\) and \(L\) on \(AB\), \(M\) on \(BC\), and \(N\) on \(CA\). Let
\[ P = AM\cap BN,\quad R = KM\cap LN,\quad S = KN\cap LM, \]and let the line \(CS\) meet \(AB\) at \(Q\). Prove that the points \(P\), \(Q\), and \(R\) are collinear.
2 replies
MathMystic33
Yesterday at 5:56 PM
AylyGayypow009
13 minutes ago
J
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Parallel lines in incircle configuration
GeorgeRP   1
N 28 minutes ago by Double07
Source: Bulgaria IMO TST 2025 P1
Let $I$ be the incenter of triangle $\triangle ABC$. Let $H_A$, $H_B$, and $H_C$ be the orthocenters of triangles $\triangle BCI$, $\triangle ACI$, and $\triangle ABI$, respectively. Prove that the lines through $H_A$, $H_B$, and $H_C$, parallel to $AI$, $BI$, and $CI$, respectively, are concurrent.
1 reply
GeorgeRP
2 hours ago
Double07
28 minutes ago
J
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Thailand geometry
EeEeRUT   2
N 37 minutes ago by Captainscrubz
Source: Thailand MO 2025 P7
Let $ABC$ be a triangle with $AB < AC$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects $BC$ at $D$. The angle bisector of $\angle BAC$ intersect $BC$ at $E$. Suppose that the perpendicular bisector of $AE$ intersect $AB, AC$ at $P,Q$, respectively. Show that $$\sqrt{\frac{BP}{CQ}} = \frac{AC \cdot BD}{AB \cdot CD}$$
2 replies
EeEeRUT
an hour ago
Captainscrubz
37 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
231 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
231 posts
GreekIdiot
#2 h  Mar 28, 2025, 8:50 AM
Y by
guys bump
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whwlqkd
102 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
102 posts
whwlqkd
#5 h  Mar 29, 2025, 3:36 AM
Y by
z^3 can be 3(mod 4)
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GreekIdiot
231 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
231 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
107 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
107 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
538 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
334 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
231 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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