Natural solutions to $4^n + 2^{n + 1} = 2^{k}$ Is there such an $n$ and $k$ that
$$4^n + 2^{n + 1} = 2^{k}$$
with $n, k \in \mathbb N$.
I wrote a program and for $n, k < 5000$ have not found a solution.
Is this possible?
 A: Hint: The left side is $2^{2n}+2^{n+1}$, which is $(2^n+1)^2-1$. When is $(x-1)(x+1)$  a power of $2$?
A: $$4^n+2^{n+1}=2^k$$ 
$$2^{2n}+2^{n+1}=2^k$$ 
$$2^{n+1}(2^{n-1}+1)=2^k$$ 
$$2^{n-1}+1=2^{k-n-1}$$
Clearly, odd number cannot be a power of $2$, hence this has no solution in $\mathbb N$ except when $n=1$ and $k=3$ as pointed out by @alex.jordan
A: It has already been mentioned in the previous answers, but I think it's worth highlighting: 4 is the square of 2, and all powers of 4 are also powers of 2. With just a little effort you should be able to verify that $4^n = 2^{2n}$.
Thus you want to solve $2^{2n} + 2^{n + 1} = 2^k$. But the only way two powers of 2 can add up to another power of 2 is if they're the same power of 2. This follows from the fact that $b + b = 2b$, and if $b$ is a power of 2, then $2b$ is the next higher power of 2. But if $b \neq c$, then $b + c \neq 2b$ nor $2c$ either, even if both $b$ and $c$ happen to be powers of 2.
So we have $2 + 2 = 4$, $4 + 4 = 8$, $8 + 8 = 16$, etc., but $2 + 4 = 2 \times 3$, $8 + 32 = 2^2 \times 5$, etc. Clearly $2^m + 2^m = 2(2^m) = 2^{m + 1}$.
This means you need to solve $2n = n + 1$, which leads to $k = 2n + 1$. There is only one possible solution in integers to $2n = n + 1$, and consequently only one solution to $2^{2n} + 2^{n + 1} = 2^{2n + 1}$, and that is $n = 1$.
