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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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[*]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
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Nine point circle + Perpendicularities
YaoAOPS   18
N 14 minutes ago by AndreiVila
Source: 2025 CTST P2
Suppose $\triangle ABC$ has $D$ as the midpoint of $BC$ and orthocenter $H$. Let $P$ be an arbitrary point on the nine point circle of $ABC$. The line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. The line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. If $M$ is the midpoint of $AQ$, show that $HX \perp DM$.
18 replies
YaoAOPS
Mar 5, 2025
AndreiVila
14 minutes ago
Inequality conjecture
RainbowNeos   0
32 minutes ago
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
0 replies
RainbowNeos
32 minutes ago
0 replies
inequality 2905
pennypc123456789   0
33 minutes ago
Consider positive real numbers \( x, y, z \) that satisfy the condition
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3.
\]Find the maximum value of the expression
\[
P = \dfrac{yz}{\sqrt[3]{3y^2z^2+ 3x^2y^2z^2+ x^2z^2 + x^2y^2}}
+ \frac{xz}{\sqrt[3]{3x^2z^2 + 3x^2y^2z^2 + x^2y^2 + y^2z^2}}
+ \frac{xy}{\sqrt[3]{3x^2y^2 + 3x^2y^2z^2 +y^2z^2 + x^2z^2}}.
\]
0 replies
pennypc123456789
33 minutes ago
0 replies
Inspired by m4thbl3nd3r
sqing   3
N 37 minutes ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
3 replies
2 viewing
sqing
Today at 3:43 AM
sqing
37 minutes ago
Inspired by qrxz17
sqing   7
N 42 minutes ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
7 replies
sqing
5 hours ago
sqing
42 minutes ago
Geometry problem
Whatisthepurposeoflife   2
N an hour ago by Whatisthepurposeoflife
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
2 replies
Whatisthepurposeoflife
Yesterday at 1:45 PM
Whatisthepurposeoflife
an hour ago
A Sequence of +1's and -1's
ike.chen   36
N an hour ago by maromex
Source: ISL 2022/C1
A $\pm 1$-sequence is a sequence of $2022$ numbers $a_1, \ldots, a_{2022},$ each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \le t_1 < \ldots < t_k \le 2022$ so that $t_{i+1} - t_i \le 2$ for all $i$, and $$\left| \sum_{i = 1}^{k} a_{t_i} \right| \ge C.$$
36 replies
ike.chen
Jul 9, 2023
maromex
an hour ago
Basic ideas in junior diophantine equations
Maths_VC   1
N an hour ago by grupyorum
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
1 reply
Maths_VC
Tuesday at 7:54 PM
grupyorum
an hour ago
Inspired by qrxz17
sqing   3
N an hour ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
3 replies
sqing
5 hours ago
sqing
an hour ago
Hardest in ARO 2008
discredit   30
N an hour ago by Phat_23000245
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
30 replies
discredit
Jun 11, 2008
Phat_23000245
an hour ago
Find the value
sqing   12
N 2 hours ago by Phat_23000245
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
12 replies
sqing
Jun 22, 2024
Phat_23000245
2 hours ago
Midpoints of arcs form a similar triangle
v_Enhance   19
N 2 hours ago by Adywastaken
Source: USA TSTST 2013, Problem 1
Let $ABC$ be a triangle and $D$, $E$, $F$ be the midpoints of arcs $BC$, $CA$, $AB$ on the circumcircle. Line $\ell_a$ passes through the feet of the perpendiculars from $A$ to $DB$ and $DC$. Line $m_a$ passes through the feet of the perpendiculars from $D$ to $AB$ and $AC$. Let $A_1$ denote the intersection of lines $\ell_a$ and $m_a$. Define points $B_1$ and $C_1$ similarly. Prove that triangle $DEF$ and $A_1B_1C_1$ are similar to each other.
19 replies
1 viewing
v_Enhance
Aug 13, 2013
Adywastaken
2 hours ago
find question
mathematical-forest   4
N 2 hours ago by GreekIdiot
Are there any contest questions that seem simple but are actually difficult? :-D
4 replies
mathematical-forest
3 hours ago
GreekIdiot
2 hours ago
4 var inequality
SunnyEvan   1
N 2 hours ago by SunnyEvan
Source: Own
Let $ x,y,z,t \in R^+ ,$ such that : $ (x+y+z+t)^2 = x+y+z+t + (x+z)(y+t) $ and $ x \geq y \geq z \geq t .$
Try to prove or disprove : $$ \frac{2 \sqrt{x+y+z+t +(x+t)(y+z)}}{x^2+y^2+z^2+t^2 +3xz+3yt+xt+yz} \geq \frac{11(x+y)(z+t)-(x+y+z+t)}{x+y+z+t +(x+z)(y+t)} $$
1 reply
SunnyEvan
Today at 7:07 AM
SunnyEvan
2 hours ago
Junior Balkan Mathematical Olympiad 2024- P3
Lukaluce   15
N May 1, 2025 by MATHS_ENTUSIAST
Source: JBMO 2024
Find all triples of positive integers $(x, y, z)$ that satisfy the equation

$$2020^x + 2^y = 2024^z.$$
Proposed by Ognjen Tešić, Serbia
15 replies
Lukaluce
Jun 27, 2024
MATHS_ENTUSIAST
May 1, 2025
Junior Balkan Mathematical Olympiad 2024- P3
G H J
Source: JBMO 2024
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Lukaluce
274 posts
#1 • 3 Y
Y by Sedro, farhad.fritl, ItsBesi
Find all triples of positive integers $(x, y, z)$ that satisfy the equation

$$2020^x + 2^y = 2024^z.$$
Proposed by Ognjen Tešić, Serbia
This post has been edited 1 time. Last edited by Lukaluce, Jun 28, 2024, 12:36 PM
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giannis2006
45 posts
#2 • 1 Y
Y by farhad.fritl
We have the following cases:
$1) y>2x$. Then we get that: $2^{2x}(505^x+2^{y-2x})=2^{3z}253^z$,so $2x=v_2(LHS)=v_2(RHS)=3z$ and hence $505^x+2^{y-2x}=253^z$, which is a contradiction by $mod 3$
$2) y<2x$. Then we get that: $2^y(2^{2x-y}505^x+1)= 2^{3z}505^z$ With the same way as case $1$ we get that $y=3z$ and hence $2^{2x-y}505^x+1=253^z$, which is a contradiction by $ mod 3$.
$3) y=2x$. Then we get that: $2^{2x}(505^x+1)=2^{3z}253^z$
$505^x+1  \equiv 2 mod 4$, and hence $2x+1=v_2(LHS)=v_2(RHS)=3z$, so equivalently we have that $505^x+1=2*253^z=2*253^{\frac {2x+1} {3}}$, which has only $x=1$ as a positive integer solution. So, in this case $(x,y,z)=(1,2,1)$ ,which is the only solution of the given equation.
This post has been edited 2 times. Last edited by giannis2006, Jun 27, 2024, 11:47 AM
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P2nisic
406 posts
#3
Y by
Consider $U_2$ we get:
$2x=3z$ contradiction since then $LHS>RHS$
$y=3z$ then $2020^x=2024^z-8^z=2016[...]$ contradiction since $7|2016$ but not $2020$
$y=2x$ we have that $2^{2x}(505^x+1)=2^{3z}*253^z$ or $505^x+1=2^{3z-2x}*253^z$
Consider $mod4$ we get that $3z-2x=1$ and then from inelyalites esily get $x=1,y=2,z=1$
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Z4ADies
64 posts
#4
Y by
First,assume that $x \geq 2 $ and $y \geq 3$.
In first case we will inspect $2x=y$.
Then,from taking both sides' power, $2x=y=3z$.
It is known that,$x>z$ but, $505^x+1=253^z$ which is contradiction.
In second case we will inspect $2x>y$.
Like first method,we get $y=3z$ then, $2x>3z$.
So, $2^{2x-y}.505^x+1 \geq 2.505^x+1 >253^z$
contradiction again.
In third case we will look $y>2x$.
From getting power both sides, we found that,$2x=3z$.$505^x+2^{y-2x}=253^z$ obviously contradicition.
So,$x=1,y=2,z=1$.
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Assassino9931
1374 posts
#5
Y by
Proposed by Serbia. Do not get negatively fooled by the classical looking statement -- the various possible approaches here can teach students a lot of important ideas!

Proposer solution, powers of 2 and size arguments

My solution, moduli and Fermat classics
This post has been edited 2 times. Last edited by Assassino9931, Jun 27, 2024, 2:33 PM
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OgnjenTesic
47 posts
#6 • 7 Y
Y by Assassino9931, oVlad, Sedro, Math_.only., ehuseyinyigit, farhad.fritl, mrtheory
Proposed by me (Ognjen Tešić, Serbia).
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Marinchoo
407 posts
#7
Y by
Taking the equation modulo $5$ yields $y$ is even, so $y=2y_1$. Now if $x\neq y_1$ we have \[2x\geq \nu_{2}(2020^x+2^y)=\min\{2x, 2y_1\}=\nu_{2}(2024^z)=3z.\]However, $2x<3z$ as $2024^z>2020^x>2024^{\frac{2}{3}x}$. Therefore $x=y_1$, and the equation becomes $4^x(505^x+1)=2024^z$. Modulo $11$ implies $x$ is odd, at which point $\nu_2(505^x+1)=\nu_2(506)=1$. Comparing the $\nu_2$'s of both sides gives $2x+1=3z$, so $x=3t+1$, $z=2t+1$ for some nonnegative integer $t$. Clearly, $t=0$ leads to the solution $(x, y, z) = (1, 2, 1)$. When $t>0$ we derive a contradiction from:
\[1>\frac{2020^{3t+1}}{2024^{2t+1}}=\frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)^t\geq \frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)>1.\]
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Davut1102
22 posts
#8
Y by
..........
This post has been edited 1 time. Last edited by Davut1102, Jul 1, 2024, 1:35 PM
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Sedro
5855 posts
#9
Y by
The only solution is $(x,y,z) = (1,2,1)$, which obviously works. We now prove it is the only one.

Claim: $y=2x$.

Proof: Take both sides of the equation modulo $3$ to get $1+2^y\equiv 2^z \pmod{3}$. Clearly, we must have $2^y \equiv 1 \pmod{3}$, so $y$ is even and $z$ is odd. Let $y = 2y_0$, for a positive integer $y_0$. Then, the given equation becomes $2020^x + 4^{y_0} = 2024^z$. Note that $v_2(2024^z) = 3z$ is always odd. Since $v_2(4)$ and $v_2(2020)$ are both even, it follows if $v_2(2020^x)\ne v_2(4^{y_0})$, then either $v_2(2020^x + 4^{y_0}) = v_2(2020^x) = 2x$ or $v_2(2020^x + 4^{y_0}) = v_2(4^{y_0})=2y_0$, neither of which are odd. Thus, $v_2(2020^x) = v_2(4^{y_0})$. Because $v_2(2020)=v_2(4)$, we must have $x=y_0$, and our claim follows.

Claim: The only possible value of $x$ is $1$.

Proof: Rewrite the given equation as $4^x(505^x + 1^x) = 2024^z$. Note that when $x=1$, $505^1+1^1=506$ is divisible by all the prime factors of $2024$, which are $2$, $11$, and $23$. If $x>1$, by Zsigmondy, there exists some $p\notin \{2,11,23\}$ that divides $505^x + 1^x$, and hence it is impossible that $505^x+1^x \mid 2024^z$. Thus, $x=1$, and we are done. $\blacksquare$
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megarnie
5611 posts
#10
Y by
The only solution is $(1,2,1)$, which clearly works.

Notice that taking the equation mod $7$ gives $4^x + 2^y \equiv 1 \pmod 7$. Since $4^3 \equiv 2^3 \equiv 1\pmod 7$, if $3$ divided either one of $x$ or $y$, then we have that one of $4^x, 2^y$ is $0\pmod 7$, which is absurd. Hence $3\nmid xy$.

Hence $\nu_2(2020^x), \nu_2(2^y)$ are both not multiples of $3$, so they cannot be equal to $ \nu_2(2024^z)$. If $\nu_2(2020^x) \ne \nu_2(2^y)$, then we would have $\nu_2(2024^z) = \nu_2(2020^x + 2^y) \in \{\nu_2(2020^x) , \nu_2(2^y)\}$, which is absurd, so $\nu_2(2020^x) = \nu_2(2^y)$, so $2x = y$. Now we have \[ 2020^x + 4^x = 2024^z \]If $x > 1$, then by Zsigmondy there exists a prime $p$ not dividing $2020^1 + 4^1 = 2024$ that divides $2020^x + 4^x$, absurd. Hence $x = 1$ must hold.
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WallyWalrus
916 posts
#11
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$\textbf{A) }$Working in modulo $10$, we obtain:
The last digit of $2020^x$ is $0$ for all $x\in\mathbb{N}$.

The last digit of $2^y$ is $\begin{cases}2,\text{ for y }\equiv 1\pmod4\\4,\text{ for y }\equiv 2\pmod4\\8,\text{ for y }\equiv 3\pmod4\\6,\text{ for y }\equiv 0\pmod4\end{cases}$

The last digit of $2024^z$ is $\begin{cases}4,\text{ for z }\equiv 1\pmod2\\6,\text{ for z }\equiv 0\pmod2\end{cases}$

Results: $y=2w,\;w\in\mathbb{N}$ and $w-z\equiv0\pmod2$.
The equation becomes:
$2020^x+4^w=2024^z$, where $w,z$ have the same parity $\quad\textbf{(1)}$.

$\textbf{B) }$Working in modulo $3$, we obtain:
$2020^x\equiv1\pmod3;\;4^w\equiv1\pmod3\Longrightarrow 2024^z\equiv2\pmod3\Longrightarrow z$ is odd number $\Longrightarrow w$ is odd number.

$\textbf{Case 1: }\min\{x,w,z\}=w;\;w\le x;\;w\le z$.
Dividing in $\textbf{(1)}$ by $4^w$ results:
$505^x\cdot 4^{x-w}+1=506^z\cdot 4^{z-w}$.
$z-w$ is an even non-negative number and results: the last digit of $506^z\cdot 4^{z-w}$ is $6$.
The last digit of $505^x\cdot 4^{x-w}$ must be $5$, hence $x=w$ and we obtain
$505^x+1=506^z\cdot 4^{z-w}$.
Working in modulo $4$ in the last relation, results:
$505^x\equiv1\pmod4\Longrightarrow 506^z\cdot 4^{z-w}\equiv2\pmod4\Longrightarrow$
$\Longrightarrow z-w=0;\;z=1\Longrightarrow x=w=z=1\Longrightarrow y=2$
and the triplet $(x,y,z)=(1,2,1)$ is solution of the equation $2020^x+2^y=2024^z$.

$\textbf{Case 2: }\min\{x,w,z\}=x;\;x<w;\;x\le z$.
Dividing in $\textbf{(1)}$ by $4^x$ results:
$505^x+4^{w-x}=506^z\cdot 4^{z-x}$.
$505^x$ is odd number; $4^{w-x}$ and $506^z\cdot 4^{z-x}$ are even numbers, hence the last equation has no solutions.

$\textbf{Case 3: }\min\{x,w,z\}=z;\;z<x;\;z<w$.
$2020^x+4^w=2024^z$.
$v_2(2024^z)=3z$, odd number (see $\textbf{B)}$).

$\textbf{case 3.1: }x<w$
$2020^x+4^w=4^x(505^x+4^{w-x})\Longrightarrow v_2(2020^x+4^w)=2x$, even number, hence the equation has no solutions.

$\textbf{case 3.2: }x>w$
$2020^x+4^w=4^w(505^x\cdot4^{x-w}+1)\Longrightarrow v_2(2020^x+4^w)=2w$, even number, hence the equation has no solutions.

$\textbf{case 3.3: }x=w$
$2020^x+4^x=2024^z$.
$z=2u+1;\;x=w>z$, where $u\in\mathbb{N}\cup\{0\}$ (see $\textbf{B)}$).
$2020^x+4^x=4^x(505^x+1)$.
$505^x+1\equiv2\pmod4\Longrightarrow v_2(505^x+1)=1\Longrightarrow$
$\Longrightarrow v_2(2020^x+4^x)=v_2(4^x(505^x+1))=2x+1=3z\Longrightarrow$
$\Longrightarrow 2x+1=6u+3\Longrightarrow x=w=3u+1;\;z=2u+1$.
$x>z\Longrightarrow u>0$.
The equation becomes:
$2020^{3u+1}+4^{3u+1}=2024^{2u+1}\Longrightarrow 505^{3u+1}+1=2\cdot253^{2u+1}\Longrightarrow$
$\Longrightarrow 505\cdot505^{3u}+1=506\cdot253^{2u}$, contradiction since
$505\cdot505^{3u}>506\cdot253^{2u},\;\forall u\in\mathbb{N}$ (proved many times in the previous posts).

$\textbf{Conclusion:}$
The equation in positive integers $2020^x+2^y=2024^z$ has the unique solution $(x,y,z)=(1,2,1)$.
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PaixiaoLover
123 posts
#13
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take mod 3 to get $1+(-1)^y=(-1)^z$. Because of this, we know y must be even and z must be odd.

Let $y=2y_1$ and $z=2z_1+1$. Prime factorizing the original equation, $2^{2x}\cdot505^x+2^{2y_1}=2^{6z_1+3}
\cdot253^{2z_1+1}$

Now considering the largest factor of 2 that divides the LHS, if $2y_1$ is the smallest factor, then $2y_1=6z_1+3$, impossible. Similarly, if 2x is the largest factor of 2, $2x=6z_1+3$ is impossible. This means $2x=2y_1$ and the factor of 2 on the LHS is $2^{2x+1}$ since the v2 of $505^{x}+1$ is at 2. (since its 0 mod 2 and not 0 mod 4). So we have $2x+1=3z_1$

Going back, we have $x=x, y=2x, z=\frac{2x+1}{3}$. since $\frac{2x+1}{3}$ is an integer, set $x=3k+1$ Plugging in and dividing by largest factor of 2 we get $505^{3k+1}+1=2\cdot253^{2k+1}.$ k=0 works, but any larger k dosent work since $505^{3k+1}+1 > 505 \cdot 505^{3k} > 506 \cdot 253^{2k}$ so the only solution is k=0, which represents $(x,y,z)=(1,2,1)$
This post has been edited 1 time. Last edited by PaixiaoLover, Jan 3, 2025, 7:56 PM
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ItsBesi
147 posts
#14 • 1 Y
Y by farhad.fritl
Here is my solution:

Answer: $(x,y,z)=(1,2,1)$

Solution:

Claim: $y-$even and $z-$odd
Proof:

By taking $\pmod 3$ we get:

$1+(-1)^y \equiv (-1)^z \pmod 3$ if $y-$ even then: $(-1)^z \equiv 0 \pmod 3$ which is a contradiction so $\boxed{y-\text{even}}$

Hence $(-1)^z \equiv -1 \pmod 3 \implies \boxed{z-\text{odd}}$ $\square$

Since $y-$ even we get that $y=2 \cdot y'$ so our equation transforms into the following:

$$2020^x+4^{y'}=2024^z$$Claim: $x=y'$
Proof: FTSOC assume $x \neq y'$

So $x \neq y' \iff 2x \neq 2y' \iff x \cdot 2 \neq y' \cdot 2 \iff x \cdot \nu_2(2020) \neq y'\cdot \nu_2(4) \iff \nu_2(2020^x) \neq \nu_2(4^{y})$

So we got that: $x \neq y' \iff \nu_2(2020^x) \neq \nu_2(4^{y})$

Hence we get that: $\nu_2(2020^x+4^{y'})= \min\{\nu_2(2020^x) , \nu_2(4^{y'}) \}= \min\{2x,2y' \}=2 \cdot \min\{x,y' \} \implies  \nu_2(2020^x+4^{y'} \}=2 \cdot \min\{x,y' \} $ $...(1)$

Also on the other hand we have that: $\nu_2(2024^z)=z \cdot \nu_2(2024)= z \cdot 3 =3 \cdot z \implies \nu_2(2024^z)=3 \cdot z$ $...(2)$

Hence we get that $3 \cdot z \stackrel{(2)}{=} \nu_2(2024^z)= \nu_2(2020^x+4^{y'} \} \stackrel{(1)}{=} 2 \cdot \min\{x,y' \} \implies 3 \cdot z=2 \cdot \min\{x,y' \}$

So $2 \mid 2 \cdot \min\{x,y' \}=3 \cdot z \implies 2 \mid 3 \cdot z \implies 2 \mid z \iff z \equiv 0 \pmod 2$ which is a contradiction because we found that $z-\text{odd}$

So our assumption is wrong hence $x=y'$ $\square$ $\implies$
$$2020^x+4^x=2024^x$$Claim: $\nu_{11}(x)=0$
Proof: FTSOC assume $\nu_{11}(x) \geq 1$

So by taking $\nu_{11}$ on both sides and using Lifting the Exponent Lemma (LTE) we get:

$1+\nu_{11}(x)=\nu_{11}(2024)+\nu_{11}(x)=\nu_{11}(2020+4)+\nu_{11}(x) \stackrel{LTE}{=}\nu_{11}(2020^x+4^x)=\nu_{11}(2024^z)=z \cdot \nu_{11}(2024)=z \implies$

$ 1+\nu_{11}(x)=z \iff z=1+\nu_{11}(x)$ Since $z- \text{odd}$ we get that $\nu_{11}(x)-\text{even}.$ Let $\nu_{11}(x)=2k  \implies z=2k+1$ also $x=11^{2k} \cdot t \implies x=121^k \cdot t$

So our equation transforms into the following:

$2020^{121^k \cdot t} + 4^{121^k \cdot t} = 2024^{2k+1}$ Now by taking $\nu_2$ we get:

$6k+3= (2k+1) \cdot 3  = (2k+1) \cdot \nu_2(2024)=\nu_2(2024^{2k+1})=\nu_2( 2020^{121^k \cdot t} + 4^{121^k \cdot t})$

$\geq \min\{\nu_2(2020^{121^k \cdot t}) , \nu_2( 4^{121^k \cdot t} ) \} =\min\{121^k \cdot t \cdot \nu_2(2020) , 121^k \cdot t \cdot \nu_2(4) \} = \min\{(121^k \cdot t \cdot 2), (121^k \cdot t \cdot 2) \}=121^k \cdot t \cdot 2$

$ \implies 6k+3 \geq 121^k \cdot t \cdot 2$ but this isn't true for $k \geq 1$

So our assumption is wrong hence $\nu_{11}(x)=0 \square$ so $z=1+\nu_{11}(x)=1 \implies \boxed{z=1}$

So $2020^x+4^x=2024$ clearly $\boxed{x=1}$ so $y=2 \cdot y' =2 \cdot x=2 \implies \boxed{y=2}$

Hence $(x,y,z)=(1,2,1)$ is the only solution $\blacksquare$
This post has been edited 4 times. Last edited by ItsBesi, May 24, 2025, 8:15 PM
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EVKV
71 posts
#15 • 3 Y
Y by Nobitasolvesproblems1979, Umudlu, Nuran2010
Claim: $y=2x$.
Proof: Analyzing $mod$ $5$ $y$ is even
let $y=2g$
now $mod$ $3$ gives $z$ is odd
$2020^{x} + 2^{2g} = 2024^{z}$ is same as $4^{x}505^{x} + 4^{g} = 2024^{z}$
Case 1: x>g
$ 2^{2g}(4^{x-g}505^{x} +1) = 2^{3z}253^{z}$
Implying $2g = 3z$ nonsense
Case 1: x<g
$ 2^{2x}(505^{x} +4^{g-x}) = 2^{3z}253^{z}$
Implying $2x = 3z$ nonsense

thus x=g

Now taking $ mod$ $ 5$ again we get $x$ is odd

$ 2^{2x}(505^{x} +1) = 2^{3z}253^{z}$
$v_2(505^{x} +1) = 1$
So, $2x+1 = 3z$
So, $x=3k+1$ (for a non-negetive integer k)
So,$z=2k+1$
Clearly (U can also induct) $2020^{3k+1} > 2024^{2k+1}$ for $k \neq 0$
thus $2020^{3k+1} +4^{3k+1} = 2020^{x} + 2^{y} > 2024^{z}$ for $k \neq 0$

Thus only solution is $(x,y,z)=(1,2,1)$
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ray66
48 posts
#16
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Taking mod 3 gives $1+(-1)^y \equiv (-1)^z \pmod 3$, so $y$ is even and $z$ is odd. Now consider $\nu_2$ of both sides. The RHS is $\nu_2(2024^z) = 3z$, so it's odd. The LHS is $\nu_2(2020^x+2^y)$, and it's odd if and only if $y=2x$. Now write the LHS as $4^x(505^x+1)$, and taking mod 4 on the inside sum gives $505^x+1\equiv 2 \pmod 4$. Now we have the relationship $z=\frac{2x+1}{3}$. We can easily check that $\boxed{(1,2,1)}$ is a solution, and we see that for $x>1$, the LHS is strictly greater than the RHS, so there are no solutions $x\ge 2$.
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MATHS_ENTUSIAST
28 posts
#17
Y by
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}

We are given the equation:
\[
2020^x + 2^y = 2024^z
\]
We aim to find natural numbers x, y, z \in \mathbb{N} that satisfy this equation.

Taking the 2-adic valuation v_2 on both sides:
\[
v_2(2020^x + 2^y) = v_2(2024^z)
\]
Note:
\[
v_2(2020) = 2 \Rightarrow v_2(2020^x) = 2x, \quad v_2(2^y) = y, \quad v_2(2024) = 4 \Rightarrow v_2(2024^z) = 4z
\]
We try different cases based on the relative sizes of 2x, y, and 4z.

\textbf{Case 1: } 2x = y < 4z

Then,
\[
2020^x + 2^y = 2^y(a + 1), \quad \text{where } a \in \mathbb{N}, \text{ odd}
\]So,
\[
v_2(2020^x + 2^y) = y = 4z \Rightarrow 2x = 4z \Rightarrow x = 2z
\]
Now reduce modulo 3:
\[
2^{2x} + 1 \equiv 0 \pmod{3} \Rightarrow 2^{2x} \equiv -1 \pmod{3}
\]
But since 2^2 \equiv 1 \pmod{3}, we get:
\[
2^{2x} \equiv 1 \pmod{3} \Rightarrow \text{Contradiction}
\]
Hence, this case is not possible.

\textbf{Case 2: } y > 2x, \quad 2x = 3z

Then:
\[
2020^x + 2^y = 2^{2x}(a + 1), \quad \text{where } a \text{ is odd}
\Rightarrow v_2 = 2x
\Rightarrow 4z = 2x \Rightarrow z = \frac{x}{2}
\Rightarrow x \text{ must be even}
\]
Try small even values of x.

\textbf{Case 3: } y < 2x, \quad y = 2x, \quad 2x + 1 = 3z

We write:
\[
2020^x + 2^{2x} = 2024^z
\Rightarrow 4^x(505^x + 1) = 2^{4z}(253^z)
\]
Note that 4^x = 2^{2x}, so:
\[
2^{2x}(505^x + 1) = 2^{4z}(253^z)
\Rightarrow 505^x + 1 = 2^{2z}(253^z)
\]
Now, 505^x + 1 must be a power of 2 times 253^z.

Try x = 1:
\[
2020^1 + 2^2 = 2020 + 4 = 2024 = 2024^1 \Rightarrow \text{Valid solution}
\]
So,
\[
x = 1,\quad y = 2x = 2,\quad 2x + 1 = 3z \Rightarrow z = \frac{2x + 1}{3} = \frac{3}{3} = 1
\]
\textbf{Now check if higher values of x can work:}

Try x = 2:
\[
2020^2 + 2^4 = 4080400 + 16 = 4080416 \\
2024^2 = 4096576 \Rightarrow \text{LHS} < \text{RHS}
\]
Try x = 3:
\[
2020^3 + 2^6 = 8204080000 + 64 = 8204080064 \\
2024^3 = 8283809024 \Rightarrow \text{LHS} < \text{RHS}
\]
As x increases, 2^y becomes insignificant, and LHS < RHS.

\textbf{Conclusion:} The only possible solution is:
\[
\boxed{(x, y, z) = (1, 2, 1)}
\]
\end{document}
This post has been edited 1 time. Last edited by MATHS_ENTUSIAST, May 1, 2025, 9:08 AM
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