Equation involving power of two

Question

I want to show that the equation $2^x - 1 = 3^y$ does not have any positive integer solutions except for $ x = 2 , y = 1$ . Is it possible to prove the assertion using binary representation of powers of three?

Answer 1

As $y$ is a positive integer, the right hand side is a multiple of $3$. But $2^x\equiv 1\pmod 3$ iff $x$ is even, say $x=2k$. Then $2^x-1=(2^k+1)(2^k-1)$ and both factors must be powers of $3$. As their difference is merely $2$, we conclude that they are $3^1=3$ and $3^0=1$.

Answer 2

One should also note that the equation $2^x-3^y=c$ has been solved already in $1935$ for all integers $c$. It is a special case of Pillai's equation $a^x-b^y=c$, and the case $(a,b)=(2,3)$ was solved by A. Herschfeld, "The equation $2^x − 3^y = d$, Bull Amer. Math. Soc., 41 (1935), p. 631." The result is as follows. For $d=1$ there is only one solution for $(x,y)$, namely $(2,1)$. There are similar cases for other small $d$, e.g., For $d=7$ it is $(4,2)$. For all $d$ bigger than $10$, or smaller than $-10$ there is at most one solution.

Answer 3

You can transform that equation into: $$ x = \log_2\left( 3^y+1 \right) $$ Now suppose $y\geq2$. It is clear that $3^y+1>8$. Thus, it is sufficient to show that $3^y+1$ is not divisible by $8$ and therefore $x$ is not an integer.

Case 1 $(y = 2k)$: $$ 3^y+1 = 3^{2k}+1 = 9^k+1 \equiv 1^k+1 = 2\qquad(\mathrm{mod }\ 8) $$ Case 2 $(y = 2k+1)$: $$ 3^y+1 = 3^{2k+1}+1 = 3\cdot9^k+1 \equiv 3\cdot1^k+1 = 4\qquad(\mathrm{mod }\ 8) $$

Thus, the only possible solution is $y=1$ and $x=2$.