Is divisible a power of 3 for a difference of powers of 2 and 3? That is, can result, this division, in an integer? $$ \frac{3^{m}}{2^{n} - 3^r} $$ where $n,m,r$ natural number.
Edit: $n>r$, $r=m+1$.
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Lely
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Trivial cases $$2^2-3^1=1, 2^3-3^2=-1$$ – lab bhattacharjee Feb 09 '14 at 09:16
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HINT:
We need $\displaystyle 2^n-3^r=\pm3^s$ where $s\le m$
$\displaystyle 2^n=3^r\pm3^s$
Case $1:$ Taking the $+$ sign, $2^n=3^r+3^s=3^{r-s}(3^s+1)$ (WLOG $r\ge s$)
As $(2,3)=1,$ we need $3^{r-s}=1\iff r=s$ and $2^n=3^r+1$
Case $2:$ Taking the $-$ sign, $2^n=3^r-3^s=3^{r-s}(3^s-1)$ (we need $r\ge s$)
Like the previous case, we must have $r=s\implies 2^n=3^r-1$
lab bhattacharjee
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What does this mean? That $\frac{3^{m}}{2^{n} -3^r}$, $n>r$, $r=m+1$ is satisfied only for $r=s$? – Lely Feb 09 '14 at 09:40
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@Lely, As explained in the answer $r=s$ is required for all $n,r\ge0$ – lab bhattacharjee Feb 09 '14 at 09:42