<s>[INST] How to solve: $\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$ $n$ is an integer variable satisfying $$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$ How can I find $n$?
 [/INST] the equation
$$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$
is equivalent to
$$2^{n-2}=\frac{n+1}{5-n}$$
because $2^{n-2}>0$ follow that $\frac{n+1}{5-n}>0,0<n<5$
we need to check for $n=1,2,3,4$
for $n=1$ we get $$2^{-1}=\frac{2}{4}=\frac{1}{2}$$
for $n=2$ we get $$2^0=\frac{3}{3}=1$$
for $n=3$ we get $$2^1=\frac{4}{2}=2$$
for $n=4$ we get $$2^2\neq\frac{5}{1}=5$$
so the integer solutions are $n=1,n=2,n=3$
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