<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] Rewriting, we have $3\cdot2^{n+1}=(n+1)(4+2^n)$
First, note that whenever $n\geq 5$, we have $(n+1)(4+2^n)>6\cdot 2^n=3\cdot2^{n+1}$
Next note that whenever $n<-1$, $(n+1)(4+2^n)<0<3\cdot2^{n+1}$
Note that $n=-1$ isn't possible because in the first equation we would divide by $0$.
Hence any solution must have $n\in[0,4]$
Now we just check our $5$ cases and find that the only solutions are $n=1$, $n=2$, and $n=3$.
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