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
11.03.2020 20:47
26.05.2024 06:34
kaede wrote:
From $E_{19}$ and $( \eighthnote )$, we finally get a contradiction.? Why??????
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 $(3,1,1)$.