SELECTION EXAMINATION

If $x=1$, then we have no solutions, while for $x=2$, we have the solution $(x,n)=(2,4)$.

Now we suppose that $x\ge 3$. The left part of the equation is even and so $n$ is even, say $n=2k$. Then the equation can be written as: $3\cdot 2^x+4=4k^2\iff 3\cdot 2^{x-2}=k^2-1$, and hence $k$ is odd. Thus the equation is written as: $3\cdot 2^{x-2}=(k-1)(k+1)$.

1st case: $k+1=3\cdot 2^a$ and $k-1=2^b$, $a,b\ge 1$, with $a+b=x-2$. Then by subtracting we get $3\cdot 2^a-2^b=2$. If $a,b\ge 2$, then the left part is divisible by 4, while the right part is not. Therefore, either $a=1$, which implies $b=2$ and $x=5$, or $b=1$, not giving solution for $a$. Hence we have $(x,n)=(5,10)$.

2nd case: $k+1=2^s$ and $k-1=3\cdot 2^t$, $s,t\ge 1$, with $s+t=x-2$. Then we have the equation $2^s-3\cdot 2^t=2$. If $s,t\ge 2$ then the left part is divisible by 4, while the right part is not. Hence, either $s=1$ or $t=1$, giving $s=3$ and $x=6$. Therefore in this case we have the solution $(x,n)=(6,14)$.