Find all triplets of positive integers $(x, y, z)$ such that $2^x+1=7^y+2^z$.
Problem
Source: Vietnam TST 2019 Day 2 P4
Tags: number theory, Diophantine equation
kaede
11.03.2020 20:47
For each positive integer $m$, let $E_{m}$ be the congruence $2^{x} +1=7^{y} +2^{z}\pmod m$.
We consider the following five cases $( 1) -( 5)$.
$( 1)$ $z=1$
We have $2^{x} =7^{y} +1$.
Assume that $x\geq 4$, then $16\mid 7^{y} +1$, impossible.
So $( x,y) =( 3,1)$ is the only solution in $( 1)$.
$( 2)$ $z=2$
We have $2^{x} =7^{y} +3$.
From $E_{8}$, we have $0\equiv 7^{y} +3\pmod 8$, impossible.
$( 3)$ $z=3$
We have $2^{x} =7^{y} +7$.
Since $7\mid \text{RHS}$, we also have $7\mid 2^{x}$, impossible.
$( 4)$ $z=4$
We have $2^{x} =7^{y} +15$.
If $x=4$, then $1=7^{y}$ , impossible.
If $x=5$, then $17=7^{y}$, impossible.
Assume that $x\geq 7$.
From $E_{128}$, we have $7^{y} +15\equiv 0\pmod{128}$, which implise $y\equiv 10\pmod{16}$.
Since $y\equiv 10\pmod 16$ and $E_{17}$, we have $2^{x} \equiv 7^{10} +15\pmod{17}$, impossible.
So $( x,y) =( 6,2)$ is the only solution in $(4)$.
$( 5)$ $z\geq 5$
From $E_{32}$, we have $2^{5} \mid 7^{y} -1$, which implise $4\mid y$.
From $4\mid y$ and $E_{5}$, we have $x\equiv z\pmod 4\cdots ( \heartsuit )$.
From $E_{7}$, we have $x\equiv 0$ and $z\equiv 1\pmod 3\ \cdots ( \clubsuit )$.
From $E_{9}$, $( \heartsuit )$, and $( \clubsuit )$,
we have $x\equiv 3,y\equiv 4,z\equiv 1\pmod 6\ \cdots ( \twonotes )$
or $x\equiv 0,y\equiv 2,z\equiv 4\pmod 6\cdots ( \eighthnote )$.
From $E_{13}$ and $( \heartsuit )$, we can verify that $( \twonotes )$ is impossible.
From $E_{19}$ and $( \eighthnote )$, we finally get a contradiction.
Hence the answer is $\boxed{( x,y,z) =( 3,1,1) ,( 6,2,4)}$.
Jjesus
26.05.2024 06:34
kaede wrote:
For each positive integer $m$, let $E_{m}$ be the congruence $2^{x} +1=7^{y} +2^{z}\pmod m$.
We consider the following five cases $( 1) -( 5)$.
$( 1)$ $z=1$
We have $2^{x} =7^{y} +1$.
Assume that $x\geq 4$, then $16\mid 7^{y} +1$, impossible.
So $( x,y) =( 3,1)$ is the only solution in $( 1)$.
$( 2)$ $z=2$
We have $2^{x} =7^{y} +3$.
From $E_{8}$, we have $0\equiv 7^{y} +3\pmod 8$, impossible.
$( 3)$ $z=3$
We have $2^{x} =7^{y} +7$.
Since $7\mid \text{RHS}$, we also have $7\mid 2^{x}$, impossible.
$( 4)$ $z=4$
We have $2^{x} =7^{y} +15$.
If $x=4$, then $1=7^{y}$ , impossible.
If $x=5$, then $17=7^{y}$, impossible.
Assume that $x\geq 7$.
From $E_{128}$, we have $7^{y} +15\equiv 0\pmod{128}$, which implise $y\equiv 10\pmod{16}$.
Since $y\equiv 10\pmod 16$ and $E_{17}$, we have $2^{x} \equiv 7^{10} +15\pmod{17}$, impossible.
So $( x,y) =( 6,2)$ is the only solution in $(4)$.
$( 5)$ $z\geq 5$
From $E_{32}$, we have $2^{5} \mid 7^{y} -1$, which implise $4\mid y$.
From $4\mid y$ and $E_{5}$, we have $x\equiv z\pmod 4\cdots ( \heartsuit )$.
From $E_{7}$, we have $x\equiv 0$ and $z\equiv 1\pmod 3\ \cdots ( \clubsuit )$.
From $E_{9}$, $( \heartsuit )$, and $( \clubsuit )$,
we have $x\equiv 3,y\equiv 4,z\equiv 1\pmod 6\ \cdots ( \twonotes )$
or $x\equiv 0,y\equiv 2,z\equiv 4\pmod 6\cdots ( \eighthnote )$.
From $E_{13}$ and $( \heartsuit )$, we can verify that $( \twonotes )$ is impossible.
From $E_{19}$ and $( \eighthnote )$, we finally get a contradiction.
Hence the answer is $\boxed{( x,y,z) =( 3,1,1) ,( 6,2,4)}$.
From $E_{19}$ and $( \eighthnote )$, we finally get a contradiction.? Why??????
Ferreirinha
29.05.2024 05:13
Let us initially support that $x>z\geq5$. Thus, $7^y\equiv 1\pmod{32}$ $\Longrightarrow$ $4|y$. Then let $y=2k$, with $k$ even. Like this: $2^x-2^z=(7^k)^2-1=(7^k+1)(7^k-1)$. Let us then prove the following lemma: Lemma: $mdc(2^{x-z}-1,7^k+1)=1$, if $x-z\geq2$. Proof: Let $mdc(2^{x-z}-1,7^k+1)=d$. Since $p$ is a prime that divides $d$, we have $p|7^k+1$ $\Longrightarrow$ $7^k\equiv -1\pmod{p}$ $\Longrightarrow$ $k$ is even, so $\left(\dfrac{-1}{p}\right)=1$ $\Longrightarrow$ $p\equiv 1\pmod{4}$. Finally, how $p|2^{x-z}-1$, $2^{x-z}\equiv 2\pmod{4}$ $\Longrightarrow$ $x-z=1,x=z+1$. Substituting, $2^z=7^y-1$, where we find no solution. If $x-z\geq2$, $p$ does not exist $\Longrightarrow$ $d=1$. Proven! Therefore, $7^k+1|2^z$ $\Longrightarrow$ $7^k+1=2^m$ $\Longrightarrow$ $v_2(2^m)=v_2(7^k+1)=3$ $ \Longrightarrow$ $m=3$. Testing, I found the solution $(x,y,z)=(6,2,4)$. If $4\geq{z}$, we will find the solution $(3,1,1)$.