Annoying 2^x-5 = 11^y

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Valentin Vornicu

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Valentin Vornicu

#1 h Jan 14, 2006, 8:25 AM • 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)

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Megus

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Megus

#2 h Jan 14, 2006, 3:46 PM • 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

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Valentin Vornicu

#3 h Jan 14, 2006, 3:53 PM • 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

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Megus

#4 h Jan 14, 2006, 4:21 PM • 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

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marko avila

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marko avila

#5 h Jan 16, 2006, 1:41 AM • 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 .

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Kalimdor

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Kalimdor

#6 h Jan 16, 2006, 1:56 AM • 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
$\Rightarrow \text{No Solution other than} (4,1) !!!$

This post has been edited 1 time. Last edited by Kalimdor, Jan 16, 2006, 7:16 AM

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Alfred

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Alfred

#7 h Jan 16, 2006, 6:13 AM • 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

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Valentin Vornicu

#8 h Jan 16, 2006, 6:47 AM • 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

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Aryabhatta

#9 h Jan 16, 2006, 8:54 AM • 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

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Valentin Vornicu

#10 h Jan 16, 2006, 11:19 AM • 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 $9\neq (2-\sqrt 5 ) (2 +\sqrt 5 ) = 4-5 = -1$

).

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

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Megus

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#11 h Jan 16, 2006, 1:17 PM • 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 $3\cdot 3 = 9 = (2-\sqrt 5 )(2+\sqrt 5) .$

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 $9\neq (2-\sqrt 5 ) (2 +\sqrt 5 ) = 4-5 = -1$

).

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

Oh, good to hear that

- but that taught me one thing - next time I'll check whether the ring I use is UFD

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Kalimdor

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#12 h Jan 16, 2006, 6:00 PM • 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

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Valentin Vornicu

#13 h Jan 16, 2006, 8:09 PM • 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

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bodom

#14 h May 19, 2007, 8:54 PM • 6 Y

Y by Adventure10, Mango247, and 4 other users

you can also solve this modulo 640(you can check it out)

but i'm too lazzy to write it down.

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scorpius119

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scorpius119

#15 h May 19, 2007, 9:14 PM • 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

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ZetaX

#16 h Jul 12, 2007, 3:39 PM • 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

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

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#17 h Aug 11, 2007, 8:29 AM • 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

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

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#18 h Aug 11, 2007, 1:10 PM • 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

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praneeth

#19 h Sep 25, 2007, 10:37 AM • 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

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ali666

#20 h Jun 12, 2008, 10:02 AM • 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

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shubham.cash

#21 h May 24, 2010, 1:59 PM • 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

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ZetaX

#22 h May 25, 2010, 9:27 PM • 2 Y

Y by Adventure10, Mango247

$6 \cdot 35 = 10 \cdot 21$ , but your conclusion is wrong.

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shubham.cash

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shubham.cash

#23 h May 26, 2010, 2:01 AM • 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

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ZetaX

#24 h May 26, 2010, 2:20 AM • 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

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shubham.cash

#25 h May 26, 2010, 2:48 AM • 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

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nnosipov

#26 h Oct 27, 2010, 6:50 PM • 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

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

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sonakshi

#27 h Dec 23, 2010, 7:28 AM • 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

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dolphinday

#28 h Oct 14, 2023, 7:18 PM • 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

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joshualiu315

#29 h Oct 28, 2023, 6:19 AM

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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

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#30 h Dec 25, 2023, 7:04 PM

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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_

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#31 h Jan 4, 2024, 8:34 PM

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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

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#32 h Feb 28, 2024, 3:53 AM

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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

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#33 h Jun 26, 2024, 9:15 PM

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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

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Pal702004

#34 h Jun 27, 2024, 1:39 PM

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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

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#35 h Aug 31, 2024, 3:44 PM

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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

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#36 h Sep 15, 2024, 6:16 PM • 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

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ryanyz

#37 h Sep 15, 2024, 6:20 PM

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No way this got bumped

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Ilikeminecraft

678 posts

Ilikeminecraft

#38 h Apr 25, 2025, 11:52 PM

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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

651 posts

Kempu33334

#39 h May 17, 2025, 4:40 PM

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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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