Greece JBMO TST

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ultralako

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ultralako

#1 h Apr 22, 2018, 4:54 PM • 3 Y

Y by Aritra12, Adventure10, Mango247

Find all positive integers $x,y,z$ with $z$ odd, which satisfy the equation:

$2018^x=100^y + 1918^z$

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Makorn

869 posts

Makorn

#2 h Apr 22, 2018, 7:04 PM • 7 Y

Y by ultralako, futurestar, Math-Ninja, KiwiYum, BobsonJoe, Adventure10, Mango247

incomplete

This post has been edited 1 time. Last edited by Makorn, Apr 22, 2018, 10:32 PM

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ultralako

10 posts

ultralako

#3 h Apr 22, 2018, 7:24 PM • 2 Y

Y by Adventure10, Mango247

Makorn wrote:

?

Indeed incomplete.

This post has been edited 1 time. Last edited by ultralako, Apr 23, 2018, 8:03 PM
Reason: .

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silouan

3952 posts

silouan

#4 h Apr 22, 2018, 8:55 PM • 2 Y

Y by Adventure10, Mango247

Makorn wrote:

Taking mod 31 gives us $3^{2a+1}\equiv{7^y+(27)^{2a+1}}$ implying $a\equiv{3}$ mod 5, a contradiction.

$a\equiv 0\pmod 5$ works also, so we don't have contradiction. Am I missing something?

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Makorn

869 posts

Makorn

#5 h Apr 22, 2018, 10:32 PM • 4 Y

Y by KiwiYum, BobsonJoe, Adventure10, Mango247

silouan wrote:

Makorn wrote:

Taking mod 31 gives us $3^{2a+1}\equiv{7^y+(27)^{2a+1}}$ implying $a\equiv{3}$ mod 5, a contradiction.

$a\equiv 0\pmod 5$ works also, so we don't have contradiction. Am I missing something?

apologies, did not realize that 0 mod 5 also worked.

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Mr.Chem-Mathy

49 posts

Mr.Chem-Mathy

#6 h Apr 23, 2018, 6:18 PM • 3 Y

Y by Iora, Adventure10, Mango247

ultralako wrote:

Find all positive integers $x,y,z$ with $z$ odd, which satisfy the equation:

$2018^x=100^y + 1918^z$

Obviously $x=y=z=1$ is a solution.
We have $100\equiv 100 (mod \;2018)$ and $1918\equiv -100(mod\; 2018)$
So $100^y\equiv 100^y (mod \;2018)$ and $1918^z\equiv -100^z(mod\; 2018)$
Clearly $100^y + (-100^z)=0 \rightarrow y=z$
Which means $2018^x=100^y + 1918^y$
$2018=2*1009$
$100^y=4^y*25^y=2^{2y}*5^{2y}$
$1918^y=$ something $*2^y$
So $2^{2y}+2^y \equiv 0 (mod \;2^x) \Leftrightarrow 2^y(2^y+1) \equiv 0 (mod;2^x)$
So finally we get $x=y$
The original equation transforms to $2018^x=100^x+1918^x$
But for $x>1$ we have $2018^x > 100^x + 1918^x$ so the only solution is $(1, 1, 1)$
I dont know how to prove the inequality above so if someone were to prove it, thanks! (Check #8 for ineq. Proof)

This post has been edited 2 times. Last edited by Mr.Chem-Mathy, Apr 23, 2018, 9:59 PM

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silouan

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silouan

#7 h Apr 23, 2018, 7:53 PM • 2 Y

Y by Adventure10, Mango247

Mr.Chem-Mathy wrote:

$2018\equiv -100(mod\; 2018)$

This is clearly not true.

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Makorn

869 posts

Makorn

#8 h Apr 23, 2018, 8:10 PM • 7 Y

Y by Mr.Chem-Mathy, KiwiYum, BobsonJoe, OlympusHero, Adventure10, Mango247, aidan0626

$2018^x=(1918+100)^x=1918^x+100^x+1918^{x-1}\cdot100+1918^{x-2}\cdot100^2+\dots\geq{1918^x+100^x}$

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WolfusA

1900 posts

WolfusA

#9 h Apr 23, 2018, 9:09 PM • 2 Y

Y by Adventure10, Mango247

with equality iff $x=1$ . Hence $y=z=1$ . Task for people with advanced visual perception.

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Mr.Chem-Mathy

49 posts

Mr.Chem-Mathy

#10 h Apr 23, 2018, 9:53 PM • 2 Y

Y by Adventure10, Mango247

silouan wrote:

Mr.Chem-Mathy wrote:

$2018\equiv -100(mod\; 2018)$

This is clearly not true.

i meant 1918, sorry.
should be fixed now

This post has been edited 1 time. Last edited by Mr.Chem-Mathy, Apr 23, 2018, 9:54 PM

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whatRthose

1792 posts

whatRthose

#11 h Apr 24, 2018, 12:17 AM • 2 Y

Y by Adventure10, Mango247

Makorn wrote:

incomplete

No, it's actually $x = \max{(2y, z)}$ Since $v_2(100^y + 1918^z) = \max{(2y, z)}$ . Also, $v_2(2018^x) = x$ , so $x = \max{(2y, z)}$

This post has been edited 1 time. Last edited by whatRthose, Apr 24, 2018, 12:17 AM

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Makorn

869 posts

Makorn

#12 h Apr 24, 2018, 12:31 AM • 4 Y

Y by KiwiYum, BobsonJoe, Adventure10, Mango247

whatRthose wrote:

Makorn wrote:

incomplete

No, it's actually $x = \max{(2y, z)}$ Since $v_2(100^y + 1918^z) = \max{(2y, z)}$ . Also, $v_2(2018^x) = x$ , so $x = \max{(2y, z)}$

Taking $y=z=1$ , we get $\nu_2(2018)=1\neq{\max{(2y,z)}}$

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franchester

1487 posts

franchester

#13 h Apr 24, 2018, 2:04 AM • 1 Y

Y by Adventure10

Makorn wrote:

incomplete

ack I am bad what do you mean by -adic?

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Makorn

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Makorn

#14 h Apr 24, 2018, 2:32 AM • 5 Y

Y by franchester, KiwiYum, BobsonJoe, Adventure10, Mango247

franchester wrote:

Makorn wrote:

incomplete

ack I am bad what do you mean by -adic?

$p$ -adic valuation of $x$ denoted by $\nu_p(x)$ is the highest power of $p$ that divides $x$ (so like $\nu_2(24)=3$ since $2^3=8$ divides 24 but $2^4=16$ doesn't)

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silouan

3952 posts

silouan

#15 h Apr 24, 2018, 9:51 AM • 2 Y

Y by Adventure10, Mango247

Mr.Chem-Mathy wrote:

Clearly $100^y + (-100^z)=0 \rightarrow y=z$

From what you said it follows that $100^y-100^z\equiv 0\pmod{2018}$ .
Why $100^y + (-100^z)=0$ ?

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rayuga

27 posts

rayuga

#16 h Apr 24, 2018, 10:37 AM • 2 Y

Y by Adventure10, Mango247

Solution

This post has been edited 9 times. Last edited by rayuga, Apr 29, 2018, 1:51 AM
Reason: Latex

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whatRthose

1792 posts

whatRthose

#18 h Apr 24, 2018, 10:04 PM • 2 Y

Y by Adventure10, Mango247

Makorn wrote:

whatRthose wrote:

Makorn wrote:

incomplete

No, it's actually $x = \max{(2y, z)}$ Since $v_2(100^y + 1918^z) = \max{(2y, z)}$ . Also, $v_2(2018^x) = x$ , so $x = \max{(2y, z)}$

Taking $y=z=1$ , we get $\nu_2(2018)=1\neq{\max{(2y,z)}}$

Oh sorry I see now

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VicKmath7

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VicKmath7

#19 h Sep 29, 2019, 12:24 PM • 1 Y

Y by Adventure10

Is there correct solution without LTE

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Illuzion

211 posts

Illuzion

#20 h Sep 30, 2019, 8:25 PM • 1 Y

Y by Adventure10

First assume $x>1$ . Taking $\mod 3$ gives that $x$ is odd.

If $x>z,$ $100^y=2018^x-1918^z=2^z(1009^x.2^{x-z}-959^z)$ and looking at the exponent of $2,$ we conclude $2y=z,$ contradiction.

If $z>x,$ analogous argument gives $2y=x,$ contradiction. Therefore $z=x$ .

Now $25^{y-1}.2^{2y-2}=2018^{x-1}+\cdots+1918^{x-1}=2^{x-1}.(1009^{x-1}+\cdots+959^{x-1})$ and the second factor is a sum of odd number of odd numbers and therefore is odd. Looking at the exponent of $2, x-1=2y-2$ . Now $25^{y-1}=1009^{x-1}+\cdots+959^{x-1}>1009^{x-1}=(1009^2)^{y-1}>25^{y-1},$ contradiction. Therefore $x=1$ and the $RHS$ is bounded and we easily find the only solution $(x, y, z) =(1, 1, 1), \blacksquare$ .

This post has been edited 1 time. Last edited by Illuzion, Sep 30, 2019, 8:26 PM
Reason: Typo

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huashiliao2020

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huashiliao2020

#22 h Mar 28, 2023, 4:24 AM

Y by

Mr.Chem-Mathy wrote:

So $2^{2y}+2^y \equiv 0 (mod \;2^x) \Leftrightarrow 2^y(2^y+1) \equiv 0 (mod;2^x)$
So finally we get $x=y$

y can be greater than x.

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Wild

18 posts

Wild

#23 h Jul 23, 2024, 3:47 PM

Y by

Mr.Chem-Mathy wrote:

ultralako wrote:

Find all positive integers $x,y,z$ with $z$ odd, which satisfy the equation:

$2018^x=100^y + 1918^z$

Obviously $x=y=z=1$ is a solution.
We have $100\equiv 100 (mod \;2018)$ and $1918\equiv -100(mod\; 2018)$
So $100^y\equiv 100^y (mod \;2018)$ and $1918^z\equiv -100^z(mod\; 2018)$
Clearly $100^y + (-100^z)=0 \rightarrow y=z$
Which means $2018^x=100^y + 1918^y$

$2018=2*1009$
$100^y=4^y*25^y=2^{2y}*5^{2y}$
$1918^y=$ something $*2^y$
So $2^{2y}+2^y \equiv 0 (mod \;2^x) \Leftrightarrow 2^y(2^y+1) \equiv 0 (mod;2^x)$
So finally we get $x=y$
The original equation transforms to $2018^x=100^x+1918^x$
But for $x>1$ we have $2018^x > 100^x + 1918^x$ so the only solution is $(1, 1, 1)$
I dont know how to prove the inequality above so if someone were to prove it, thanks! (Check #8 for ineq. Proof)

The last inequality is Fermat's big theorem

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Nuran2010

49 posts

Nuran2010

#24 h Dec 7, 2024, 5:52 PM • 2 Y

Y by Omerking, TunarHasanzade

Mr.Chem-Mathy wrote:

ultralako wrote:

Find all positive integers $x,y,z$ with $z$ odd, which satisfy the equation:

$2018^x=100^y + 1918^z$

Obviously $x=y=z=1$ is a solution.
We have $100\equiv 100 (mod \;2018)$ and $1918\equiv -100(mod\; 2018)$
So $100^y\equiv 100^y (mod \;2018)$ and $1918^z\equiv -100^z(mod\; 2018)$
Clearly $100^y + (-100^z)=0 \rightarrow y=z$

Don't we conclude y and z have same parity here?For example if y is even,then z is even too.However,we are given z odd,so y will be odd as well.I did not catch why do we have y=z

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lksb

163 posts

lksb

#25 h Dec 7, 2024, 6:40 PM • 1 Y

Y by AkosS

$\nu_2(2018^x)=x=\min\{2y, z\}$ Assume $x=2y$ , by mod 3, we have that $(-1)^x\equiv2\pmod{3}\iff 2\nmid x=2y, \text{Abs!}$
Therefore, $\boxed{x=z}$
From that, the equation becomes $2018^x-1918^x=100^y$ Let $x,y>1$
By Zsigmondy's Theorem, $\exists p\in\mathbb{P} s.t. p\mid 2018^x-1918^x \wedge p\nmid 2018-1918=100, \text{Abs!}$
Therefore, $\boxed{(x,y,z)=(1,1,1)}$ is the only solution!

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Davut1102

16 posts

Davut1102

#26 h Dec 7, 2024, 7:31 PM

Y by

Makorn wrote:

incomplete

I think(and hope) I can finish this with LTE(I hope there is no calculation and language error):

If x,y>2,because x is odd and 5|2018-1918=100,LTE says that $2y=v_5(100^y)=v_5(2018^x - 1918^x) = v_5(100)+v_2(x)$ and it is clear that x>y.
But then $2018^x$ is smaller than $1918^x + 100^x$ ,absurd and we are done.