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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?

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  • $\begingroup$ The LHS is $2^{n+1}(2^{n-1}-1)$ hence the odd number $2^{n-1}-1$ would divide $2^k$. Ergo? $\endgroup$ – Did Feb 7 '15 at 17:44
  • $\begingroup$ If your program started with $n=1$ and didn't find a solution, you need to work on your programming skills. $\endgroup$ – Joffan Feb 7 '15 at 17:47
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    $\begingroup$ Think about what the expression $4^n + 2^{n+1} = 2^k$ looks like in binary. $\endgroup$ – DanielV Feb 7 '15 at 17:53
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$$4^n+2^{n+1}=2^k$$ $$2^{2n}+2^{n+1}=2^k$$ $$2^{n+1}(2^{n-1}+1)=2^k$$ $$2^{n-1}+1=2^{k-n-1}$$
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

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    $\begingroup$ The left side is even if $n=1$ though, and there is one solution $(n,k)=(1,3)$. Also, why mention $\mathbb{R}$? There are many solutions in $\mathbb{R}$. OP is looking for natural number solutions. $\endgroup$ – alex.jordan Feb 7 '15 at 17:48
  • $\begingroup$ @alex.jordan Thanks for pointing that out $\endgroup$ – AvZ Feb 7 '15 at 17:49
  • $\begingroup$ Sure thing. I didn't down vote by the way. Your answer is pretty good overall. $\endgroup$ – alex.jordan Feb 7 '15 at 17:50
  • $\begingroup$ As pointed out by alex.jordan, this has infinitely many solutions in $\mathbb R$, so please change that to $\mathbb N$. $\endgroup$ – user26486 Feb 7 '15 at 19:02
  • $\begingroup$ It was suppose to mean natural solutions in $\mathbb R$, but okay. Now it's fine. $\endgroup$ – AvZ Feb 7 '15 at 19:08
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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 $2$?

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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 $n = 1$.

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    $\begingroup$ "the only way two powers of 2 can add up to another power of 2 is if they're the same power of 2" Isn't this more or less assuming the answer? $\endgroup$ – Did Feb 8 '15 at 8:53
  • $\begingroup$ @Did Maybe. I guess I should try to explain how I arrived at the conclusion, however obvious it may seem to you and I. I have edited my answer, hopefully the explanation is clear enough. $\endgroup$ – Robert Soupe Feb 8 '15 at 17:46

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