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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
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0 replies
jwelsh
Jul 1, 2025
0 replies
abc=1
hctb00   11
N 3 minutes ago by SunnyEvan
Source: unknown
$a,b,c>0,abc=1$,prove:\[\frac{1}{a^2+b+1}+\frac{1}{b^2+c+1}+\frac{1}{c^2+a+1}\le1\]
11 replies
hctb00
Aug 16, 2014
SunnyEvan
3 minutes ago
maximizing score
KevinYang2.71   3
N 10 minutes ago by Assassino9931
Source: ISL 2024 C1
Let $n$ be a positive integer. A class of $n$ students run $n$ races, in each of which they are ranked with no draws. A student is eligible for a rating $(a,\,b)$ for positive integers $a$ and $b$ if they come in the top $b$ places in at least $a$ of the races. Their final score is the maximum possible value of $a-b$ across all ratings for which they are eligible.

Find the maximum possible sum of all the scores of the $n$ students.
3 replies
KevinYang2.71
Today at 3:00 AM
Assassino9931
10 minutes ago
orang NT
KevinYang2.71   14
N 12 minutes ago by star-1ord
Source: ISL 2024 N1
Find all positive integers $n$ with the following property: for all positive divisors $d$ of $n$, we have $d+1\mid n$ or $d+1$ is prime.
14 replies
KevinYang2.71
Today at 3:00 AM
star-1ord
12 minutes ago
Inspired by old results
sqing   0
13 minutes ago
Source: Own
Let $ a,b> 0,a+b+1=ab .$ Prove that
$$   \frac{1}{a^2+ab+1}+ \frac{1}{b^2+ab+1}  \leq \frac{2(7-4\sqrt 2)}{17}$$$$   \frac{a}{a^2+ 2 ab+1}+ \frac{b}{b^2+  2ab+1} \leq \frac{2\sqrt 2-1}{7}$$Let $ a,b> 0,a+b+2=ab .$ Prove that
$$   \frac{1}{a^2+2ab+1}+ \frac{1}{b^2+2ab+1}  \leq  \frac{2(13-6\sqrt 3)}{61}$$$$   \frac{a}{a^2+ ab+1}+ \frac{b}{b^2+ ab+1} \leq  \frac{2(5\sqrt 3-3)}{33}$$
0 replies
1 viewing
sqing
13 minutes ago
0 replies
product of 5 consec integers [Original Problem]
aaa12345   0
2 hours ago
Find the number of positive integer divisors of \[S=\sum_{k=1}^{100}k(k+1)(k+2)(k+3)(k+4).\]Answer
Solution
0 replies
aaa12345
2 hours ago
0 replies
bishops attacking each other [2025 POMO1]
aaa12345   0
2 hours ago
In chess, a bishop attacks any piece that is in the same diagonal. I am placing two bishops on a $100 \times 100$ chessboard on different squares. Let $N$ be the number of possible placements there are in which they are attacking each other. Find the first three digits of $N$.

Answer
Solution
Source: 2025 Penchick Online Mathematical Olympiad Qualifying 1 II. #10
0 replies
aaa12345
2 hours ago
0 replies
Tangent at the orthocenter
Tofa7a._.36   1
N 2 hours ago by aaravdodhia
Let $ABC$ be a triangle. Consider the points $D,E,F$ as the feet of the altitudes from $A,B,C,$ respectively, and $H$ its orthocenter which we suppose is the midpoint of $CF$. Let $M$ be the midpoint of $BC$, $N$ be the midpoint of $BE$, and $X=(AN)\cap(MF).$
Prove that $\angle HXM=90^\circ$.
1 reply
Tofa7a._.36
Today at 1:17 AM
aaravdodhia
2 hours ago
IOQM P12 2024
SomeonecoolLovesMaths   10
N 2 hours ago by Lunatic_Lunar7986
Consider a square $ABCD$ of side length $16$. Let $E,F$ be points on $CD$ such that $CE = EF = FD$. Let the line $BF$ and $AE$ meet in $M$. The area of $\bigtriangleup MAB$ is:
10 replies
SomeonecoolLovesMaths
Sep 8, 2024
Lunatic_Lunar7986
2 hours ago
Nt and chains
Tofa7a._.36   1
N 2 hours ago by alexheinis
Let \( a_0, a_1, \ldots, a_n \) be positive divisors of the number \( 2024^{2025} \) such that:

\(\bullet\) \( a_0 < a_1 < a_2 < \cdots < a_n \)

\(\bullet\) \( a_0 \mid a_1,\, a_1 \mid a_2,\, \ldots,\, a_{n-1} \mid a_n \)

Find the largest possible value of the positive integer \( n \).
1 reply
Tofa7a._.36
Today at 1:13 AM
alexheinis
2 hours ago
combination
Doanh   1
N 2 hours ago by alexheinis
\textbf{Problem 14.} There are \( m \) girls and \( n \) boys. It is known that for any two boys and any two girls, there exists at least one boy--girl pair who do not know each other (mutually). Prove that the total number of boy--girl pairs who know each other (regardless of order) does not exceed
\[
m + \frac{n(n - 1)}{2}.
\]
1 reply
Doanh
Today at 6:20 AM
alexheinis
2 hours ago
Inequalities
sqing   29
N Today at 6:07 AM by sqing
Let $ x,y,z $ be real numbers such that $ x^2+2y^2+z^2+xy+yz+zx=1 $.Prove that$$-\sqrt{\frac 75}\leq  x+y+z   \leq  \sqrt{\frac 75}$$
29 replies
sqing
Aug 31, 2024
sqing
Today at 6:07 AM
Inequalities
sqing   12
N Today at 6:05 AM by sqing
Let $ a,b,c $ be real numbers . Prove that
$$- \frac{64(9+2\sqrt{21})}{9} \leq \frac {(ab-4)(bc-4)(ca-4) } {(a^2+a +1)(b^2+b +1)(c^2+c +1)}\leq \frac{16}{9}$$$$- \frac{8(436+79\sqrt{31})}{27} \leq  \frac {(ab-5)(bc-5)(ca-5) } {(a^2+a +1)(b^2+b +1)(c^2+c +1)}\leq \frac{25}{12}$$
12 replies
sqing
Jul 6, 2025
sqing
Today at 6:05 AM
Inequalities
sqing   18
N Today at 6:05 AM by sqing
Suppose that $x$ and $y$ are nonzero real numbers such that $\left(x + \frac{1}{y} \right) \left(y + \frac{1}{x} \right) = 5$. Prove that
$$\left( x^2-y^2 + \frac{1}{y^2} \right) \left(y^2-x^2 + \frac{1}{x^2} \right)\leq  \frac{7+3\sqrt{5}}{2}$$$$\left( x^3 -y^3+ \frac{1}{y^3} \right) \left(y^3 -x^3+ \frac{1}{x^3} \right)\leq  9+4\sqrt{5}$$
18 replies
sqing
Jul 2, 2025
sqing
Today at 6:05 AM
[20th PMO Area Stage: I. 5]
reilynso   2
N Today at 6:04 AM by P0tat0b0y
Let $f(x)=\sqrt{4 \sin^4 x - \sin ^2 x \cos ^2 x + 4 \cos^4 x}$ for any $x \in \mathbb{R}$. Let $M$ and $m$ be the maximum and minimum values of $f$, respectively. Find the product of $M$ and $m$.

Answer

Solution
2 replies
reilynso
Jul 12, 2025
P0tat0b0y
Today at 6:04 AM
Annoying 2^x-5 = 11^y
Valentin Vornicu   38
N May 17, 2025 by Kempu33334
Find all positive integer solutions to $2^x - 5 = 11^y$.

Comment (some ideas)
38 replies
Valentin Vornicu
Jan 14, 2006
Kempu33334
May 17, 2025
Annoying 2^x-5 = 11^y
G H J
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Valentin Vornicu
7301 posts
#1 • 5 Y
Y by narutomath96, Adventure10, Mango247, and 2 other users
Find all positive integer solutions to $2^x - 5 = 11^y$.

Comment (some ideas)
Z K Y
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Megus
1198 posts
#2 • 6 Y
Y by narutomath96, Adventure10, Mango247, and 3 other users
Make a substitution $x \to 4x$ (as Valentin said it is easy to prove - look modulo $5$). So our equation will take form $(4^x-\sqrt{5})(4^x+\sqrt{5})=11^y$. Suppose that there is a prime $p$ which divides both $4^x-\sqrt{5}$ and $4^x+\sqrt{5}$. Then $p|2 \cdot 4^x$ but also $p|11^y$ - contradiction. Hence $(4^x-\sqrt{5},4^x+\sqrt{5})=1$ and both are $y$-th powers. Because $11^y=(4+\sqrt{5})^y(4-\sqrt{5})^y$ we must have $4^x+\sqrt{5}=(4+\sqrt{5})^y$. Now using binomial expansion we get (looking at $\sqrt{5}$) $1=\sum \binom{y}{2k+1}5^k4^{y-(2k+1)}$ and from it it is easy to see that we must have $y=x=1$ (because else - right side would be larger).

Hence the only solution is $(4,1)$.

Does it make any sense ? :)
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Valentin Vornicu
7301 posts
#3 • 4 Y
Y by Jupiter123, Adventure10, Mango247, and 1 other user
Megus wrote:
Does it make any sense ? :)
No it doesn't. Let me tell you why: because $\mathbb{Z}[\sqrt 5]$ is not an UFD. For example \[ 3\cdot 3 = 9 = (2-\sqrt 5 )(2+\sqrt 5) . \]
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Megus
1198 posts
#4 • 2 Y
Y by Adventure10, Mango247
Valentin Vornicu wrote:
Megus wrote:
Does it make any sense ? :)
No it doesn't. Let me tell you why: because $\mathbb{Z}[\sqrt 5]$ is not an UFD. For example \[ 3\cdot 3 = 9 = (2-\sqrt 5 )(2+\sqrt 5) .  \]

Ooops :blush:
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marko avila
521 posts
#5 • 4 Y
Y by Adventure10, Mango247, and 2 other users
i remember seeing these type of diophantine equations being solved using algebraic number theory , and if congruences dont work then inequalities is my biggest bet. but thats just my opinion . :lol:
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Kalimdor
163 posts
#6 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Could we try this:
Appearently, $(4,1)$ is a solution
Assume that there's another solution $(x_1, y_1)$ other than this one, and of course bigger than this one.(I mean, $x_1>4, y_1>1$)

$\Rightarrow 2^{x_1}-2^4 = 11^{y_1}-11$
$\Rightarrow (2^{x_1-4}-1)(2^4) = 11(11^{y_1-1}-1)$
$\Rightarrow 11|2^{x_1-4}-1$ Simply because 11 is a prime number
$\Rightarrow \text{No Solution other than} (4,1) !!!$ :P
This post has been edited 1 time. Last edited by Kalimdor, Jan 16, 2006, 7:16 AM
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Alfred
36 posts
#7 • 1 Y
Y by Adventure10
Unfortunately $11|2^{x_1-4}-1$ has numerous solutions, just let $x_1-4=9k+1$ ($k$ integer of course).

Disclaimer: I am tired, and I might have done something wrong. But at least the last time I checked, $11|1024-1$ etc.
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Valentin Vornicu
7301 posts
#8 • 2 Y
Y by Adventure10, Mango247
Of course you cannot solve this modulo 11, since there is a solution! You have to try at least modulo 121 or 32. I tried both these, and modulo 64 also. Nothing :)
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Aryabhatta
122 posts
#9 • 3 Y
Y by Adventure10, Mango247, and 1 other user
I guess the Unique factorization can be used, but we must use $Z[\sqrt{11}]$

As you yourself have shown, $x = 4m$ and $y = 4k+1$.

Let $2^{2m} = a$ and $11^{2k} = b$. Then we have that

$a^2 - 11b^2 = 5$
i.e

$(a-b\sqrt{11})(a+b\sqrt{11}) = 5$.

Now we can apply the fact that $Z[\sqrt{11}]$ is a UFD. Which is true, I think.
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Valentin Vornicu
7301 posts
#10 • 6 Y
Y by narutomath96, Adventure10, Hexagrammum16, Mango247, and 2 other users
Valentin Vornicu wrote:
Megus wrote:
Does it make any sense ? :)
No it doesn't. Let me tell you why: because $\mathbb{Z}[\sqrt 5]$ is not an UFD. For example \[ 3\cdot 3 = 9 = (2-\sqrt 5 )(2+\sqrt 5) .  \]
Now why didn't anyone notice the HUGE abberation I have wrote above? :)

I ment $\mathbb{Z}[\sqrt{ - 5} ] = \mathbb{Z}[i\sqrt 5]$ is not an UFD (indeed $9\neq (2-\sqrt 5 ) (2 +\sqrt 5 ) = 4-5 = -1$ :D).

Actually both $\mathbb{Z}[\sqrt 5]$ and $\mathbb{Z}[\sqrt{11}]$ are Euclidean rings, thus UFDs also, so Megus' solution above is correct :) :blush:
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Megus
1198 posts
#11 • 2 Y
Y by Adventure10, Mango247
Valentin Vornicu wrote:
Valentin Vornicu wrote:
Megus wrote:
Does it make any sense ? :)
No it doesn't. Let me tell you why: because $\mathbb{Z}[\sqrt 5]$ is not an UFD. For example \[ 3\cdot 3 = 9 = (2-\sqrt 5 )(2+\sqrt 5) .  \]
Now why didn't anyone notice the HUGE abberation I have wrote above? :)

I ment $\mathbb{Z}[\sqrt{ - 5} ] = \mathbb{Z}[i\sqrt 5]$ is not an UFD (indeed $9\neq (2-\sqrt 5 ) (2 +\sqrt 5 ) = 4-5 = -1$ :D).

Actually both $\mathbb{Z}[\sqrt 5]$ and $\mathbb{Z}[\sqrt{11}]$ are Euclidean rings, thus UFDs also, so Megus' solution above is correct :) :blush:

Oh, good to hear that :) - but that taught me one thing - next time I'll check whether the ring I use is UFD :D
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Kalimdor
163 posts
#12 • 2 Y
Y by Adventure10, Mango247
I checked again and found those interesting stuffs
1. the endings of $11^y +5$ are always "$26$","$46$", "$66$", "$86$".(dunno how to prove this, maybe there's some exceptions)
2. the endings of $16^{x/4}$ are always "$16$","$36$", "$56$", "$96$".
(Appearently, x is a multiple of 4)
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Valentin Vornicu
7301 posts
#13 • 2 Y
Y by Adventure10, Mango247
Kalimdor wrote:
1. the endings of $11^y +5$ are always "$26$","$46$", "$66$", "$86$".(dunno how to prove this, maybe there's some exceptions)
That cannot be correct, as $11^y + 5 \equiv 16 \pmod {100}$ when $y \equiv 1 \pmod {\varphi (100) = 40}$.
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bodom
123 posts
#14 • 6 Y
Y by Adventure10, Mango247, and 4 other users
you can also solve this modulo 640(you can check it out) :wink: :D but i'm too lazzy to write it down. :blush:
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scorpius119
1677 posts
#15 • 5 Y
Y by hyperbolictangent, Adventure10, rstenetbg, Hexagrammum16, Mango247
Actually, you can use mods 64 and 17.
Click to reveal hidden text
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ZetaX
7579 posts
#16 • 2 Y
Y by Adventure10, Mango247
Valentin Vornicu wrote:
Actually both $ \mathbb{Z}[\sqrt 5]$ and $ \mathbb{Z}[\sqrt{11}]$ are Euclidean rings, thus UFDs also, so Megus' solution above is correct :) :blush:
Just a small correction: $ \mathbb{Z}[\sqrt 5]$ is in fact not an UFD (e.g. $ 2 \cdot 2 = (\sqrt 5-1)(\sqrt 5+1)$). But $ \mathbb{Z}[\frac{1+\sqrt 5}2]$ is, implying the same way of solution.
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Tomaths
56 posts
#17 • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
Valentin Vornicu wrote:
Actually both $ \mathbb{Z}[\sqrt 5]$ and $ \mathbb{Z}[\sqrt{11}]$ are Euclidean rings, thus UFDs also, so Megus' solution above is correct :) :blush:
Just a small correction: $ \mathbb{Z}[\sqrt 5]$ is in fact not an UFD (e.g. $ 2 \cdot 2 = (\sqrt 5-1)(\sqrt 5+1)$). But $ \mathbb{Z}[\frac{1+\sqrt 5}2]$ is, implying the same way of solution.

is there like a list of all $ \mathbb{Z}[\sqrt{d}]$ which is UFD?

could some one provide a link

thanks
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ZetaX
7579 posts
#18 • 2 Y
Y by Adventure10, Mango247
Instead of $ \mathbb Z [\sqrt d ]$, it's more convenient to consider the so-called integral closure. For $ d \not\equiv 1 \mod 4$ and squarefree, this is the same. But if $ d \equiv 1 \mod 4$, you consider $ \mathbb Z [ \frac{1+\sqrt d}2]$. And if $ d$ is not squarefree, you first divide $ d$ through the largest square dividing $ d$.
If you don't want the integral closure, a bit more work is required (you consider the "conductor" [correct english word¿]).

It's conjectured that there are infinitely many such $ d$. A complete list is known for $ d<0$.
But instead of looking for a list of UFD's, you may look for a list of class numbers (class number $ 1$ $ \iff$ UFD). See http://mathworld.wolfram.com/ClassNumber.html for example.
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praneeth
82 posts
#19 • 2 Y
Y by Adventure10, Mango247
2^4*(2^(x-4)-1)=11*(11^(y-1)-1)
11^(y-1) always ends in 1. So, 11^(y-1)-1 is divisible by 10.
But LHS is not divisible by 10 except for x=4.
So, x=4, y=1 is the only possible solution.
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ali666
352 posts
#20 • 2 Y
Y by Adventure10, Mango247
praneeth wrote:
2^4*(2^(x-4)-1)=11*(11^(y-1)-1)
11^(y-1) always ends in 1. So, 11^(y-1)-1 is divisible by 10.
But LHS is not divisible by 10 except for x=4.
So, x=4, y=1 is the only possible solution.
you are wrong,the LHS is divisible by $ 10$ for all $ x=4k$
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shubham.cash
16 posts
#21 • 2 Y
Y by Adventure10, Mango247
Though this is really old, but I just feel like posting my solution.

Replace 5 by 11-6

You get,
2 ( 2^(x-1) + 3 ) = 11(11^(y-1) + 1 )

Clearly, both factors on both sides are coprime.

Thus,
2 = 11^(y-1) + 1

Which gives y=1
Similarly x=4
Thus, (4,1) is the solution
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ZetaX
7579 posts
#22 • 2 Y
Y by Adventure10, Mango247
$6 \cdot 35 = 10 \cdot 21$, but your conclusion is wrong.
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shubham.cash
16 posts
#23 • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
$6 \cdot 35 = 10 \cdot 21$, but your conclusion is wrong.

I think the counterexample you put forward, isn't proving me wrong. My equation satisfies two conditions:

1. Both factors on either sides are coprime. (Satisfied here too).
2. 2 doesn't divide 11, But we notice that 6 divides 10 AND 6 divides 21.

Thus, you can't apply that here.

Am I right?
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ZetaX
7579 posts
#24 • 2 Y
Y by Adventure10, Mango247
I have some doubts about $6$ dividing $10$ or $21$ ;)
And you could use the example $2 \cdot 3 = 1 \cdot 6$, and still cannot conclude that $6=2$.
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shubham.cash
16 posts
#25 • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
I have some doubts about $6$ dividing $10$ or $21$ ;)
And you could use the example $2 \cdot 3 = 1 \cdot 6$, and still cannot conclude that $6=2$.

Oops. Sorry. My bad. I was excited about the fact that I had found the solution in a simpler way. :)
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nnosipov
245 posts
#26 • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
Valentin Vornicu wrote:
Actually both $ \mathbb{Z}[\sqrt 5]$ and $ \mathbb{Z}[\sqrt{11}]$ are Euclidean rings, thus UFDs also, so Megus' solution above is correct :) :blush:
Just a small correction: $ \mathbb{Z}[\sqrt 5]$ is in fact not an UFD (e.g. $ 2 \cdot 2 = (\sqrt 5-1)(\sqrt 5+1)$). But $ \mathbb{Z}[\frac{1+\sqrt 5}2]$ is, implying the same way of solution.

In $ \mathbb{Z}[\frac{1+\sqrt 5}2]$ there exist infinitely many of units (invertible elements of this UFD). So, we need to consider the case $4^x+\sqrt{5}=\varepsilon(4+\sqrt{5})^y$ where $\varepsilon$ is arbitrary unit. Megus' solution is incomplete. Ok?
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sonakshi
16 posts
#27 • 2 Y
Y by Adventure10, Mango247
@nnosipov: It is hard to read what you wrote!
[mod edit: nnosipov's post has been fixed]
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dolphinday
1335 posts
#28 • 2 Y
Y by happypi31415, endless_abyss
We will claim that the only solution is $(x, y) = (4, 1)$, which is clearly true as $16 - 5 = 11$.

Clearly $x \le 3$ does not work.

We can prove that it is impossible for $x \geq 5$.

Taking$\pmod{5}$ gets us $2^x \equiv 1\pmod{5}$, so $4|x$.

Taking$\pmod{32}$, and $x \geq 5$ gets us $27 \equiv 11^y\pmod{32}$.

So, $y \equiv 5\pmod{8}$ for $x \geq 5$.

Letting $x = 4c$, and then taking$\pmod{17}:$

$(-1)^{c} - 5 \equiv 11^y\pmod{17}$. The RHS is equal to $11$ or $13\pmod{17}$.
The LHS is equal to $10$ or $15\pmod{17}$, so by contradiction, $x$ cannot $\geq 5.$

Hence, the only solution is $(x, y) = (4, 1)$.
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joshualiu315
2535 posts
#29
Y by
The only pair that works is $(x,y) = \boxed{(4,1)}$. This can be easily checked to work so we will show it is the only one.

To begin, manually check that there are no other solutions for $x \le 4$, so assume $x>4$ for the remainder of the solution. Modulo $5$ gives $4 \mid x \implies x =4z$ for $z>1$. Thus, modulo $64$ gives

\[11^y \equiv 59 \pmod{64} \implies y \equiv 13 \pmod{16}.\]
Finally, use mod $17$ to give

\[16^z - 5 \equiv 11^y \equiv 11^{13} \equiv 7 \pmod{17}\]
which clearly has no solutions.
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Shreyasharma
687 posts
#30
Y by
Take modulo $16$ first to see that we need, $$11^{y-1} \equiv 1 \pmod{16}$$for $x \geq 4$. Then we must have $y \equiv 1 \pmod{4}$. Similarly modulo $5$ gives $4 \mid x$. Substituting this becomes, $$(2^4)^{x'} - 5 = 11 \cdot (11^4)^{y'}$$Then taking modulo $17$ we find, $$(-1)^{x'} - 5 \equiv 11\cdot 4^{y'} \pmod{17}$$Then this rearranges to $$ 11 \cdot 4^{y'} - (-1)^{x'} \equiv 12 \pmod{17}$$Also taking modulo $64$ we have $$11^y \equiv 59 \pmod{64}$$which requires $y \equiv 13 \pmod{16}$. Now it is easy to see that there is no why satisfying our modulo $17$ equation so we are done. For $x \leq 4$ our only solutions is $\boxed{(4, 1)}$.
This post has been edited 1 time. Last edited by Shreyasharma, Dec 25, 2023, 7:04 PM
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peppapig_
286 posts
#31
Y by
I claim that $(4,1)$ is the only solution. Clearly it works. Plugging in all $0<x<4$ also gives that no $x<4$ works.

Now, taking both sides mod $5$, note that the RHS is always $1$ mod $5$ no matter what $y$ is, meaning that the LHS must also be $1$ mod $5$. Using our knowledge of orders, we deduce that $2^x$ is $1$ mod $5$ if and only if $4\mid x$. So let $x=4a$ for some integer $a$. We now have that
\[16^a-5=11^y.\]Now for the sake of contradiction, assume that there's a solution $(x,y)$ with $x>4$. Taking the LHS mod $32$, since $a$ is a positive integer and $a>1$, we have that $32\mid 16^a$. Therefore the LHS is $27$ mod $32$, meaning that the RHS must also be $27$ mod $32$.

Listing the powers of $11$ mod $32$, we get
\[11, 25, 19, 17, 27, 9, 3, 1, \dots,\]which gives us that $y$ must be $5$ mod $8$. Now we take the equation mod $17$. Since $16$ is $-1$ mod $17$, we have that the LHS mod $17$ must be either $13$ or $11$. However, listing the powers of $11$ mod $17$, we get
\[11, 2, 5, 4, 10, 8, 3, 16, 6, 15, 12, 13, 7, 9, 14, 1, \dots,\]which gives us that whenever we have $y$ as $5$ mod $8$, the RHS must be $7$ or $10$ mod $17$, a contradiction to the $11$ or $13$. Therefore there are no solutions where $x>4$, meaning that $(4,1)$ is the only solution, finishing the problem.
This post has been edited 1 time. Last edited by peppapig_, Jan 4, 2024, 8:34 PM
Reason: \dots
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shendrew7
816 posts
#32
Y by
The only solution where $1 \leq x \leq 5$ is $\boxed{(x,y)=(4,1)}$. Otherwise, we subtract 11 from both sides to get
\[2^x-16 = 11^y-11 \implies 16(2^{x-4}-1) = 11(11^{y-1}-1).\]
Taking modulo 5 then modulo 64, we find $x \equiv 0 \pmod 4$ and $y \equiv 13 \pmod{16}$. However, substituting these values back in, we find the equation has no solutions modulo 17 when $x \ge 6$. $\blacksquare$
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SenorSloth
37 posts
#33
Y by
We claim that the only solution is $(4,1)$, which clearly works. We can quickly check that having $x<4$ or $y=0$ fails, so for what follows, we will assume $x>4$ and $y>1$ and show that this is impossible.

We must have $2^x\equiv 1\pmod{5}$ so $4\mid x$. We also require that $2^x\equiv 5 \pmod{11}$, so we must have $x\equiv 4\pmod{10}$. Putting these together we have that $x\equiv 4\pmod{20}$.

Now we rewrite the equation as $2^x-2^4= 11^y-11$. Since the left side is $16\pmod{32}$, we require $11^y\equiv 27\pmod{32}$. After computing a few powers of $11\pmod{32}$, we can determine that we must have $y\equiv 5\pmod{8}$. Since $x\equiv 4\pmod{20}$, we know that $2^{20}-1$ divides the left side. By noting that $41$ is a factor of $1025=2^{10}+1$, we know that $41$ must divide the left side and thus also the right side, so we need $11^{y-1}=1\pmod{41}$. Using the fact that $11^2\equiv -2\pmod{41}$ to simplify our computations, we can determine that the order of $11\pmod{41}$ is $40$ and thus $y\equiv 1\pmod{40}$. However, we also require $y\equiv 5\pmod{8}$, contradiction.
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Pal702004
543 posts
#34
Y by
No solutions for $x \ge 6$

$2^x=11^y+5$

$64 \mid 11^y+5\Longrightarrow y=16n+13$

$11^{16n+13}+5 \equiv 11^{13}+5 \equiv 12 \pmod{17}$

$2^x \not \equiv 12 \pmod{17}$
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bebebe
1002 posts
#35
Y by
Taking mod $5$ gives $2^x \equiv 1 \pmod{5},$ so $x \equiv 0 \pmod{4}.$ Letting $x=4a$ gives $$16^a-5=11^y.$$Taking mod $11$ gives $$5^a \equiv 5 \pmod{11},$$so $10|a-1.$ Write $a=10b+1.$ If $b=0,$ we get the solution $(4,1)$. Now assume $b \ge 1.$ Taking mod $32$ gives $$-5 \equiv 11^y \pmod{32}.$$After experimenting some values, we see $11^5 \equiv -5 \pmod{32}.$ Since $11$ has a order of $8$ mod $32$ (we can check $11^8 \equiv 1 \pmod{32}$ but $11^4 \ne 1 \pmod{32}$), let $y=5+8t.$ Taking mod $17$ gives $$(-1)^{10b+1} - 5 \equiv 11^{5+8t} \pmod{17},$$which simplifies to (note that $\phi(17)=16=2\cdot 8$) $$11 \equiv 11^{5, 13} \pmod{17}.$$Either possibility gives a contradiction, so the only solution is $\boxed{(4,1)}.$
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happypi31415
777 posts
#36 • 1 Y
Y by dolphinday
We claim that the only solution is $(4,1)$. It can be manually checked that all solutions with $x<4$ won't work, so from now we will assume that $x>4$. First, taking mod 5 gives us that $2^x \equiv 1 \pmod{5}$ so $4|x$. Therefore, we can rewrite as $16^x-5=11^y$. Now, taking $\pmod{32}$ gives us $$11^y \equiv 27 \pmod{32} \implies (8+3)^y \equiv 27 \pmod{32}.$$This means that $3^y \equiv 3 \pmod{8}$ by binomial theorem, so $y$ must be odd. Therefore, we can let $y=2a+1$.


Now, combining all of our results so far, we have $16^x-5=11 \cdot 121^a$. Taking mod 17 now gives us $$(-1)^x-5=11 \cdot 2^y \pmod{17}.$$Therefore, we have $11 \cdot 2^a = \pm 1$. When $1$ is positive, there are no solutions, but when $1$ is negative, we have $2^a \equiv 1 \pmod{17} \implies 4|a$. Therefore, now, we can write $y=8b+1$.


However, note that the order of $11 \pmod{32}$ is $8$, so we have that $$11^y \equiv 11^{8b+1} \equiv 11 \pmod{32},$$contradicting the earlier $11^y=27 \pmod{32}$, so we are done.
This post has been edited 1 time. Last edited by happypi31415, Sep 15, 2024, 6:17 PM
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ryanyz
128 posts
#37
Y by
No way this got bumped
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Ilikeminecraft
731 posts
#38
Y by
I claim that the only answer is $(x, y) = (4, 1).$ This is clearly valid. It is obvious that any $x < 6, x \neq 4$ doesn't work. Hence, assume that $x\geq 6.$

By taking modulo $64,$ we have that $y\equiv5\pmod{16}.$ By taking modulo $17,$ we get that $2^x \equiv5 + 11^5 \equiv 15\pmod{17}.$ This implies that $x\equiv5\pmod8\implies x\equiv1\pmod2.$

Subtract $11$ from both sides to get $2^x - 16 = 11^y - 11.$ Now, take $v_{11}$ on both sides. We get that $v_{11}(2^x - 16) = 1.$ Clearly, we have that $10 \mid x - 4,$ which means that $x\equiv0\pmod2.$

Thus, a contradiction.
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Kempu33334
786 posts
#39
Y by
We try all $x$ up to $5$, finding that the only pair is $(4,1)$. Next, we take mod $32$. For $x \ge 6$, we find that the $2^x$ term drops, leaving \[11^y \equiv 59 \pmod{64}.\]We can solve this, getting that $y \equiv 13 \pmod{16}$. Next, we take mod $17$ (\textit{For my reference: FLT gives $p-1$}). This gives that $2^x - 5\equiv 11^y \pmod{17} \equiv 11^{y\pmod{16}} \pmod{17}$ by FLT. This means we only need to try $y = 13$, which doesn't give a valid solution (for $x \ge 6$). Thus, our answer is $\boxed{(4,1)}$.
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