Make square number

Number Theory Level 3

Find the sum of all integers $n$ which make $2^{8}+2^{11}+2^{n}$ a perfect square.

(1981 HMO modification)

The answer is 12.

2 solutions

Brian Charlesworth
Nov 10, 2014

Note first that $2^{8} + 2^{11} = 2304 = 48^{2}$ .

So we are looking for integers $n$ such that

$48^{2} + 2^{n} = m^{2} \Longrightarrow 2^{n} = m^{2} - 48^{2} = (m - 48)(m + 48)$

for some integer $m$ .

By the Fundamental Theorem of Arithmetic we must have both of $|m - 48|$ and $|m + 48|$ be powers of $2$ . The only such powers that differ by $2 * 48 = 96$ are $2^{5} = 32$ and $2^{7} = 128$ . As a result, we must have $2^{n} = 2^{5} * 2^{7} = 2^{12}$ , thus giving us $\boxed{12}$ as the solution.

(Note there are two possible values for $m$ , namely $80$ and $-80$ , but both of course lead to the same value of $n$ .)

Yes sir did it same way , was very exicted to post the solution,upvoted

U Z - 6 years, 7 months ago

Thanks, Megh. It's a nice problem, and it makes me think of further problems. For example, find all integers $n$ such that $2^{8} + 2^{11} + 2^{n}$ is a perfect cube, (I'm not even sure if there are any). We could also play around with the exponents and generalize, e.g., finding all triples $(a,b,c)$ such that $2^{a} + 2^{b} + 2^{c}$ is a perfect square, (cube, etc.), (with possible restrictions on $a,b,c$ ).

As always, one good problem leads to another. :)

Brian Charlesworth - 6 years, 7 months ago

$2^{a} + 2.2^{b-1} + 2^{c}$

Perfect square , therefore $a = 2m , c = 2n$

$b - 1 = m + n$

$b + 1 = m + n + 1$

Cube , $2^{a} + 2^{b} + 2^{c} = 2^{3n} = 8^{n}$

$a = 3x , c = 3y$

$2^{b} = 3.2^{3(x + y)}(2^{x} + 2^{y})$

but $2^{b}$ will give always a multiple of 2. therefore its a contradiction.

By pascal's triangle , it should contain 4 terms , but here there are 3 , therefore not possible. @brian charlesworth

U Z - 6 years, 7 months ago

@U Z – @Megh Choksi Good analysis. I see how letting $a = 2m, c = 2n$ and hence $b = m + n + 1$ leads to a perfect square, but I'm not sure how $1, (m + n + 1), b$ will form an A.P.. Perhaps there is a typo there. :)

As for perfect cubes, how about

$2^{1} + 2^{1} + 2^{2} = 2^{3} = 8$ or $2^{4} + 2^{4} + 2^{5} = 4^{3} = 64$ ?

In general, for $a = 3n - 2, n \ge 1$ , we have

$2^{a} + 2^{a} + 2^{a+1} = 2^{a} * 2^{2} = 2^{a+2} = 2^{3n} = (2^{n})^{3}$ .

Brian Charlesworth - 6 years, 7 months ago

@Brian Charlesworth – Really good observation Sir

Thank you .Sir you are great . One day I will post here only for power of 4 , 5 ,............ , .Thank you for being humble.

U Z - 6 years, 7 months ago

@U Z – I appreciate the compliments, Megh. I'll keep working on the generalized problem as well. :)

Brian Charlesworth - 6 years, 7 months ago

Trevor Arashiro
Nov 10, 2014

Because this is a perfect square, it must have the representation $(2^a+2^b)^2$ where a,b are integers.

Upon squaring, we get $2^{2a}+2^{a+b+1}+2^{2b}$

We can observe that out of the three terms, only one can have an odd power, thus n must be even.

In this case, we let a=4 so our first term is $2^8$ and our second term is $2^{5+b}\Rightarrow 2^{5+b}=2^{11} ~ \therefore b=6.$ since n=2b, $\boxed{n=12}$

Nice generalization, Trevor. My solution was more specific to this problem, so I prefer your solution. :)

Brian Charlesworth - 6 years, 7 months ago

Because this is a perfect square, it must have the representation $2^a + 2^b$ where a,b are integers.

I don't understand how you got that. Can you elaborate?

Siddhartha Srivastava - 6 years, 7 months ago

Sorry. That should be squared, I just fixed it

Trevor Arashiro - 6 years, 7 months ago

Make square number

The answer is 12.

2 solutions

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