Solving $2^x+2^y+1=3^z$ in integers

Question

I have reasons to believe that there should be an elementary, relatively simple way to find all solutions of the equation in the title in positive integers $x>y$ and $z$. Any ideas?

Answer 1

I think I was eventually able to recover (?) the elementary solution mentioned in the literature. I will show below that the only solutions in positive integers $x,y,z$ are $x=4,y=6,z=4$, $x=6,y=4,z=4$, and $x=y=z=2$.

Reducing $2^x+2^y+1=3^z$ modulo $3$, we conclude that both $x$ and $y$ are even; say, $x=2a$ and $y=2b$, leading to $4^a+4^b+1=3^z$. Reducing now modulo $4$ we see that $z$ is even, too, and the substitution $z=2c$ gives $4^a+4^b+1=9^c$.

Assume without loss of generality that $a\le b$. If $a=1$, then $4^b+5=9^c$ whence $b=1$ and therefore also $c=1$, as otherwise reducing modulo $16$ we would get $9^c\equiv 5\pmod{16}$, which is impossible (the order of $9$ modulo $16$ is $2$). Thus $a\ge 2$, and reducing $4^a+4^b+1=9^c$ modulo $16$ we obtain $9^c\equiv 1\pmod{16}$, whence $c$ is even; we write $c=2u$ to get $4^a+4^b+1=81^u$.

We know that $a\ge 2$. If, indeed, $a=2$, then $4^b+17=81^u$, whence $(9^u-2^b)(9^u+2^b)=17$, leading to $u=1,b=3$; that is, $x=z=4$, $y=6$.

We are thus left with the case where $b\ge a\ge 3$. Reducing $4^a+4^b+1=81^u$ modulo $64$ and observing that the order of $81$ modulo $64$ is $4$, we conclude that $u$ is divisible by $4$, whence $4^a+4^b+1=3^{16v}$, $u=4v$. Reducing then modulo $85$ (which is a divisor of $3^{16}-1$), we get $4^a+4^b\equiv 0\pmod{85}$. Dividing through by $4^a$ gives $4^{b-a}\equiv -1\pmod{85}$. However, there is no power of $4$ congruent to $-1$ modulo $85$, as one can easily verify (the order of $4$ modulo $85$ is $4$). This shows that there are no solutions with $a,b\ge 3$.

Answer 2

Just a bunch of considerations for now.

Let we define the Hamming weight $H$ of $n$ as the number of bits "1" in the binary representation of $n$. We have, for instance, $H(7)=3$, $H(15)=4$, $H(2^m)=1$. Moreover, let $a_k=3^k-1$.
We have $a_{k+1}=(2+1)a_k+2$.

Claim. For every $k\geq 5$, $H(a_k)\geq 3. \tag{1}$

If we prove the above claim, we obviously have that $(x,y,z)=(6,4,4)$ is the only solution of the given equation. We may consider that: $$\begin{array}{rcr} a_4 &=& 1010000_{2}\\ a_5&=&11110010_2\\ a_6&=&1011011000_2\\ a_7&=&100010001010_2\\ a_8&=&1100110100000_2 \\ a_9&=&100110011100010_2\end{array}$$ and that if $a_k\pmod{16}\not\in\{0,1,2,4,8\}$ then $H(a_k)$ is $\geq 3$ for sure.
The same conclusion holds if $a_k\pmod{32}\not\in\{0,1,2,4,8,16\}$.

If $k$ has to be a multiple of $4$ as stated by W-t-P in the comments and proved by Jyrki Lahtonen,
it makes sense to set $b_k=a_{4k}$ and prove

Claim. For every $k\geq 2$, $H(b_k)\geq 3. \tag{2}$

maybe through $b_{k+1}=(2^6+2^4+2^0) b_k + (2^6+2^4)$ and/or

$$ \log_2(3) \approx \frac{1623}{1024} = 1.1001010111_2.$$

The question is indeed settled by a result of Dubitskas, giving that the distance of $\left(\frac{3}{2}\right)^k$ from the closest integer is $\geq 2^{-ck}$ with $c\approx 0.793$. Such a result in diophantine approximations follows from the work of Beukers, Thue and Siegel (the hypergeometric method).

Answer 3

The only solutions are given by a comment to equation (8.036) in the article http://dx.doi.org/10.2140/pjm.1982.101.263 by Brenner and Foster (Pac. J. Math 101 (1982) no 2, p 263-301).

Answer 4

@W-t-P MINOR NIT: There is also the trivial solution $x$=0, $y$=0, $z$=1. And since the original problem says 'solution in integers', not 'solution in natural numbers', one must also note that there are no solutions containing negative numbers (because either one gets an equality between an integer and a fraction [on either side], or after one clears fractions one gets an equality between a multiple of a power of three and an exact power of two).