i prove that for a large $N$,such that $m,n>N$,the equation of $3^n-2^m=l$ has no solution
we rewrite above eqution to form:
$3^n=(2^{m/2}-il^{1/2})(2^{m/2}-il^{1/2})$
if $gcd(2^{m/2}-il^{1/2}),(2^{m/2}-il^{1/2})=d$
similar to $z[i]$,$NORM(2^{m/2}+-il^{1/2})=2^m+l=3^n$ and $NORM(2^{{m/2}+1})=2^{m+2}$,then $d=1$
so $2^{m/2}+il^{1/2}=(a+ib)^n$ and $2^{m/2}-il^{1/2}=(a-ib)^n$,that $3=a^2+b^2$, we can write:
$2^{m/2}+il^{1/2}=(a+ib)^n=3^{n/2}(cos(nx)+isin(nx))$,and also
$2^{m/2}-il^{1/2}=(a+ib)^n=3^{n/2}(cos(nx)-isin(nx))$ and $tan(x)=b/a$
then $cos(nx)=2^{m/2}/3^{n/2}$ and $sin(nx)=l^{1/2}/3^{n/2}$
so $tan(nx)=l^{1/2}/2^{m/2}$by lagrange's theorem,there is a $c$,such that $c$ is between $a_1$ and $b_1$ and
$f'(c)=f(b_1)-f(a_1)/(b_1-a_1)$,then $(1+tan^{2}(c))=tan(nx)/nx$ ,or $tan(nx)>nx$
therfore $x<(l^{1/2}/2^{m/2})n$<(l^{1/2})/(2^{m/2}n)$
for $m,n>N$, $x$ is almost 0,then$b$ is almost zero,since$2^{m/2}+il^{1/2}=(a+ib)^n$,$l$ should be almost zero and this is contradiction,so for large $N$ this equation has no solution for a fix$l$
