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If $n$ is a natural number, and $2^{10} + 2^{13} + 2^n$ is a perfect square, what is the value of $n$?

I've attempted to factor out $2^{10}$ and got $2^{10}(1 + 2^3 + 2^{n-10})$. How can I move further?

player3236
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Cyh1368
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    Guess that $(1+2^3+2^{n-10})=(1+2^a)^2$. Solve for $a$. ā€“ lulu Nov 15 '20 at 13:05
  • Very similar to this. AFAICT a factor of $2^2$ is the only difference. As I happened to answer that one, I won't vote to close as a duplicate as the first voter. If somebody else feels strongly about it, do ping me. ā€“ Jyrki Lahtonen Nov 16 '20 at 12:37

3 Answers3

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Complete the square:

$2^{10}+2^{13}+2^n=(2^5)^2+2\cdot2^5\cdot2^{n/2}+2^n=(2^5+2^{n/2})^2. $

$13=1+5+n/2$.

Can you take it from here?

J. W. Tanner
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  • But could it be the square of an integer of some other form for some other $n$? (Answer: No, see my answer to the linked question). ā€“ Jyrki Lahtonen Nov 16 '20 at 12:39
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$$(a+b)^2=a^2+2ab+b^2$$

$$2^{10}+2^{13}+2^n=(2^5)^2+2\times2^{12}+(2^7)^2=(2^5)^2+2\times2^{5+7}+(2^7)^2=(2^5+2^7)^2$$

Hence $n=14$

Lion Heart
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We can factor out the $2^{10}$ first:

$$2^{10}(1+2^3+2^{nāˆ’10})$$

We can simplify $2^3 = 2\cdot{2}\cdot{2} = 8$. Also, because $2^{10}$ is a perfect square, we can remove it from the equation.

$$1+8+2^{n-10}=9+2^{n-10}$$

Now we can add a new variable, $m$, where $m$ can be any positive integer. We can rewrite our equation as:

$$9+2^{n-10}=m^2$$

We can now test values for $n-10$. If we do this, we'll find that $9 + 2^4 = 9+16=25=5^2$.

So $n-10=4$, and $n=14.$

Mathemagician314
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