Can anyone prove this assertion? Or at least suggest a method of attack? It has come up in my research.
There do not exist $a,b$ and $c$ such that$$ (2a-1)(2^{(b+c)}-3^c )=2^b-1 $$where $a>0,b>1,c>1$ and $a,b,c ∈ Z$
This question came up as I was comparing 2 types of sums: $S1=x+(3/2)x+(3/2)^2x+...+(3/2)^cx, S2=y+2y+2^2y+...+2^by$ to see if they could ever equal one another, given specific constraints on the relationship between x and y - specifically that $y=2x−1$ and $x=2^c(2a−1) $
Let me rewrite the letters for your variables due to my long-time practice.
I usually write $N$ for $c$ , $S$ for $c+b$ such that $S = \lceil N \cdot \log_2(3) \rceil$ and $B$ for $b$.
Also, to simplify I write $k$ for $2a-1$, not forgetting $k$ must be odd.
I refer also to a formula of G. Rhin, cited in J. Simons [Si,07], for a lower bound of $S \log 2 - N \log 3$ depending on $N$.
So we start with your formula, simply rewritten in notation, and then adapting for application of Rhin's inequality: $$ \begin{array} {rl} k(2^S-3^N)&=2^B - 1 \\ k( (2^S/3^N)-1)&=(2^B - 1)/3^N \\ 2^S/3^N &= 1 + (2^B - 1)/k/3^N \end{array} \tag 1$$ Logarithmizing and using the Mercatorseries on the rhs gives us $$\small {S \ln2 - N \ln3 = (2^B - 1)/k/3^N -1/2((2^B - 1)/k/3^N)^2+ 1/3((2^B - 1)/k/3^N)^3 \pm \ldots }$$
Cancelling of trailing small terms on the rhs gives now an in-equality $$ S \ln2 - N \ln3 \lt (2^B - 1)/k/3^N \tag 2 $$
By G. Rhin we have due to J. Simons (formula slightly rewritten to be better memorizable): $$\frac1{457}\frac1{N^{13.3}} \lt S \ln2 - N \ln3 \tag 3 $$ So we can conclude $$\begin{array} {rl} \frac1{457}\frac1{N^{13.3}} &\lt S \ln2 - N \ln3 &\lt (2^B - 1)/k/3^N \\ \frac1{457}\frac1{N^{13.3}} &\lt (2^B - 1)/k/3^N \end{array} \tag 4 $$ Now taking logarithms again gives $$\small \begin{array} {rl} -6.13 - 13.3 \ln N &\lt \ln (2^B - 1) - \ln k - N \ln 3 \\ N \ln 3 - 13.3 \ln N &\lt 6.13 + B \ln 2 -(1/2^B + 1/2/4^B + ..) - \ln k \\ N \ln 3 - B \ln 2 - 13.3 \ln N &\lt 6.13 -(1/2^B + 1/2/4^B + ..) - \ln k \\ \end{array} \tag 5$$ Here obviously the lhs is increasing with increasing $N$ while the rhs stays roughly constant (or even decreases if you increase $k$) so we can solve for the equality-condition given $N$ and $\small{B=\lceil N \cdot (\log_2 (3) -1)\rceil }$ and some assumed $k$. Say $k=3$ then at $\small {N=95.05}$ the lhs grows over the rhs.
So for $N \gt 95$ there is no solution.
The cases $N \le 95$ can be done one by one finding that no other solution exists for $N>2$
Now check for other odd $k>3$.
So you are done.
Remark: I think this is easier than the Simon's exposition because I do not need to refer to the theory of continued fractions here.
[Si,07] John L Simons: On the (non)-existence of m-cycles for generalized Syracuse sequences
$ \qquad \qquad $(2007) (online update of earlier article)
[Rh,87] G. Rhin: Aproximants de Padé et mesures d’irrationalité.
$ \qquad \qquad $Progress in Mathematics 71, (1987), pp. 155-164
$ \qquad \qquad $(Reference supplied by [Si,07])