Find all positive integers $x,y,z$ with $z$ odd, which satisfy the equation: $$2018^x=100^y + 1918^z$$
Problem
Source: Greece JBMO TST Problem 4
Tags: number theory, Exponential equation
Makorn
22.04.2018 22:04
2-adic valuation on both sides gives $x=\min{(2y,z)}$ (unless $2y=z$ but $z$ is odd so not possible) and taking mod 3 gives $2^x\equiv{1+1=2}$ so $x$ is odd, implying $x=z$
Obviously $(1,1,1)$ works.
Now transform the condition to $2018^x-1918^x=100^y$. The left side factors out to $(100)(2018^{x-1}+2018^{x-2}\cdot1918+2018^{x-3}\cdot1918^2+\dots+1918^{x-1})$. Canceling a $100$ on both sides and taking mod 100 gives $x(18^{x-1})\equiv{0}$ mod 100 and so $x\equiv{25}$ mod 50.
ultralako
22.04.2018 22:24
Makorn wrote:
2-adic valuation on both sides gives $x=\min{(2y,z)}$ (unless $2y=z$ but $z$ is odd so not possible) and taking mod 3 gives $2^x\equiv{1+1=2}$ so $x$ is odd, implying $x=z$
Obviously $(1,1,1)$ works. Now substitute $\frac{x-1}{2}=\frac{z-1}{2}=a$ and get $2018^{2a+1}=100^y+1918^{2a+1}$ for positive integer $a$. But taking mod 11 gives us $5^{2a+1}\equiv{1+4^{2a+1}}$ implying $a\equiv{0}$ mod 5. Taking mod 31 gives us $3^{2a+1}\equiv{7^y+(27)^{2a+1}}$ implying $a\equiv{3}$ mod 5, a contradiction. Thus $(1,1,1)$ is the only pair.
Indeed incomplete.
silouan
22.04.2018 23:55
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?
Makorn
23.04.2018 01:32
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.
Mr.Chem-Mathy
23.04.2018 21:18
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)
silouan
23.04.2018 22:53
Mr.Chem-Mathy wrote: $ 2018\equiv -100(mod\; 2018)$ This is clearly not true.
Makorn
23.04.2018 23:10
$2018^x=(1918+100)^x=1918^x+100^x+1918^{x-1}\cdot100+1918^{x-2}\cdot100^2+\dots\geq{1918^x+100^x}$
WolfusA
24.04.2018 00:09
with equality iff $x=1$. Hence $y=z=1$. Task for people with advanced visual perception.
Mr.Chem-Mathy
24.04.2018 00:53
silouan wrote: Mr.Chem-Mathy wrote: $ 2018\equiv -100(mod\; 2018)$ This is clearly not true. i meant 1918, sorry. should be fixed now
whatRthose
24.04.2018 03:17
Makorn wrote:
2-adic valuation on both sides gives $x=\min{(2y,z)}$ (unless $2y=z$ but $z$ is odd so not possible) and taking mod 3 gives $2^x\equiv{1+1=2}$ so $x$ is odd, implying $x=z$
Obviously $(1,1,1)$ works.
Now transform the condition to $2018^x-1918^x=100^y$. The left side factors out to $(100)(2018^{x-1}+2018^{x-2}\cdot1918+2018^{x-3}\cdot1918^2+\dots+1918^{x-1})$. Canceling a $100$ on both sides and taking mod 100 gives $x(18^{x-1})\equiv{0}$ mod 100 and so $x\equiv{25}$ mod 50.
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 $x = \max{(2y, z)}$
Makorn
24.04.2018 03:31
whatRthose wrote: Makorn wrote:
2-adic valuation on both sides gives $x=\min{(2y,z)}$ (unless $2y=z$ but $z$ is odd so not possible) and taking mod 3 gives $2^x\equiv{1+1=2}$ so $x$ is odd, implying $x=z$
Obviously $(1,1,1)$ works.
Now transform the condition to $2018^x-1918^x=100^y$. The left side factors out to $(100)(2018^{x-1}+2018^{x-2}\cdot1918+2018^{x-3}\cdot1918^2+\dots+1918^{x-1})$. Canceling a $100$ on both sides and taking mod 100 gives $x(18^{x-1})\equiv{0}$ mod 100 and so $x\equiv{25}$ mod 50.
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 $\nu_2(2018)=1\neq{\max{(2y,z)}}$
franchester
24.04.2018 05:04
Makorn wrote:
2-adic valuation on both sides gives $x=\min{(2y,z)}$ (unless $2y=z$ but $z$ is odd so not possible) and taking mod 3 gives $2^x\equiv{1+1=2}$ so $x$ is odd, implying $x=z$
Obviously $(1,1,1)$ works.
Now transform the condition to $2018^x-1918^x=100^y$. The left side factors out to $(100)(2018^{x-1}+2018^{x-2}\cdot1918+2018^{x-3}\cdot1918^2+\dots+1918^{x-1})$. Canceling a $100$ on both sides and taking mod 100 gives $x(18^{x-1})\equiv{0}$ mod 100 and so $x\equiv{25}$ mod 50.
ack I am bad what do you mean by -adic?
Makorn
24.04.2018 05:32
franchester wrote: Makorn wrote:
2-adic valuation on both sides gives $x=\min{(2y,z)}$ (unless $2y=z$ but $z$ is odd so not possible) and taking mod 3 gives $2^x\equiv{1+1=2}$ so $x$ is odd, implying $x=z$
Obviously $(1,1,1)$ works.
Now transform the condition to $2018^x-1918^x=100^y$. The left side factors out to $(100)(2018^{x-1}+2018^{x-2}\cdot1918+2018^{x-3}\cdot1918^2+\dots+1918^{x-1})$. Canceling a $100$ on both sides and taking mod 100 gives $x(18^{x-1})\equiv{0}$ mod 100 and so $x\equiv{25}$ mod 50.
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)
silouan
24.04.2018 12:51
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 $100^y + (-100^z)=0$?
rayuga
24.04.2018 13:37
We can factored the expressions and use cases.
We can see, it is given that $$2^{x}*1009^{x}=2^{2y}*5^{2y}+2^{z}*959^{z}$$Checking $\pmod{3}$ on both side we see that
$2018^{x} \equiv 2^{x}\equiv 2\pmod{3}$ thus
$$\boxed{x \equiv 1\pmod{2}}$$
Case 1: Let $x>z$ then we can arrive the equation $$2^{x-z}*1009^{x}-959^{z}=2^{2y-z}*5^{2y}$$Now $RHS$ is an integer so $LHS$ must be an integer thus as $x$ is not even then
$2y-z\geq 1$ as and $x-z\geq 1$ thus considering $\pmod{2}$ we arrives a contradiction since the above expression modulo $2$ implies $959^{z}\equiv 0\pmod{2}$
Thus $x\leq z$
Case 2: Similar to Case 1 $x<z$ implies $$1009^{x}-2^{z-x}*959^{z}=2^{2y-x}*5^{2y}$$As $RHS$ is an integer then so $LHS$ thus $2y-x\geq 1$ and $z-x\geq 1$ hence $\pmod{2}$ implies $1009^{z}$ is even a contradiction thus $\boxed{x=z}$
Case 3: Plug $x=z$ and assume $2y-x>1$ then consider $\pmod{4}$ to the resulted equation then we have $$1 \equiv 1009^{x} \equiv 959^{x}\equiv (-1)^{x} \pmod{4}$$which implies $x$ is even a contradiction thus $2y-x=1$
Finally $x=2y-1=z$ which is equivalent to the equation $$1009^{2y-1}-959^{2y-1}=2*5^{2y}$$we note that $5|1009-959$ hence via LTE we see that $$2y=v_{5}(2*5^{2y})=v_{5}(1009^{2y-1}-959^{2y-1})=2+v_{5}(2y-1)$$thus $$2y-1\equiv 0\pmod{5^{2y-2}}$$which is clearly false as $5^{2y-2}>2y-1$ for every $y>1$ in integer thus $y=1$
Hence $x=y=z=1$ is the only solution
$\square$
Note: However if $2y-z$ in case 1 is negative we can establish a contradiction $\pmod{2}$ . Similar argument can be considered for $2y-x$ in case 2 and that yields, we can take each of those expressions as positive in our solution if necessary.
whatRthose
25.04.2018 01:04
Makorn wrote: whatRthose wrote: Makorn wrote:
2-adic valuation on both sides gives $x=\min{(2y,z)}$ (unless $2y=z$ but $z$ is odd so not possible) and taking mod 3 gives $2^x\equiv{1+1=2}$ so $x$ is odd, implying $x=z$
Obviously $(1,1,1)$ works.
Now transform the condition to $2018^x-1918^x=100^y$. The left side factors out to $(100)(2018^{x-1}+2018^{x-2}\cdot1918+2018^{x-3}\cdot1918^2+\dots+1918^{x-1})$. Canceling a $100$ on both sides and taking mod 100 gives $x(18^{x-1})\equiv{0}$ mod 100 and so $x\equiv{25}$ mod 50.
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
VicKmath7
29.09.2019 15:24
Is there correct solution without LTE
Illuzion
30.09.2019 23:25
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 $(x, y, z) =(1, 1, 1), \blacksquare$.
huashiliao2020
28.03.2023 07:24
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.
Wild
23.07.2024 18:47
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
Nuran2010
07.12.2024 20:52
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
lksb
07.12.2024 21:40
$$\nu_2(2018^x)=x=\min\{2y, z\}$$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!
Davut1102
07.12.2024 22:31
Makorn wrote:
2-adic valuation on both sides gives $x=\min{(2y,z)}$ (unless $2y=z$ but $z$ is odd so not possible) and taking mod 3 gives $2^x\equiv{1+1=2}$ so $x$ is odd, implying $x=z$
Obviously $(1,1,1)$ works.
Now transform the condition to $2018^x-1918^x=100^y$. The left side factors out to $(100)(2018^{x-1}+2018^{x-2}\cdot1918+2018^{x-3}\cdot1918^2+\dots+1918^{x-1})$. Canceling a $100$ on both sides and taking mod 100 gives $x(18^{x-1})\equiv{0}$ mod 100 and so $x\equiv{25}$ mod 50.
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.