3 added 2 characters in body

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$2 deleted 1 characters in body i prove that for a large$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$(1+tan^{2}(nc))=tan(nx)/nx$(1+tan^{2}(c))=tan(nx)/nx$ ,or $tan(nx)>nx$

therfore $x<(l^{1/2}/2^{m/2})n$

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$l$

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 $(1+tan^{2}(nc))=tan(nx)/nx$ ,or $tan(nx)>nx$

therfore $x<(l^{1/2}/2^{m/2})n$

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$l$