Find the largest positive integer $n$ such that $2^{2016}+2^{6012}+2^n$ is a perfect square number.

The answer is 10006.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

I assumed that since n was the largest positive integer, it would be greater than 2016.

$2^{2016}+2^{6012}+2^n$ can be written as $2^{2016}.(2^{n-2016} + 2^{3996}+1)$

Now $2^{2016}$ is already a square number, so for the second bracket we can use the following expansion:

$(2^{m}+1)^{2}= 2^{2m}+2^{m}+1$

Therefore:

$2m=n-2016$

$m+1=3996$

Since doing the other substitution, i.e:

$2m=3996$

$m+1=n-2016$

yields a smaller n, i.e: n= 4015.

Therefore,On solving n comes out as 10006

4 Helpful
0 Interesting
0 Brilliant
0 Confused

Nice assumption and method. Your solution is better than other's observatory solutions.

Priyanshu Mishra
- 5 years, 5 months ago

Log in to reply

'More good' is grammatically incorrect

William Isoroku
- 5 years, 5 months ago

Log in to reply

Sorry, i have changed "more good" to "better" which is correct grammatically. Thanks for identifying mistakes.

Priyanshu Mishra
- 5 years, 5 months ago

It's possible to prove that your answer is the largest. What are your thoughts? Give it a try.

Daniel Liu
- 5 years, 5 months ago

We can rewrite the equation as the following:

$(2^{1008}+2^{3006})^2=2^{2016}+2^{3006}+2^n$ in which case $n=4015$

Or another possibility is:

$(2^{1008}+2^{\frac{n}{2}})^2=2^{2016}+2^{6012}+2^{n}$ . In this case, $n=10006$

Clearly the second case yields a way larger value of $n$

2 Helpful
0 Interesting
0 Brilliant
0 Confused

0 Helpful
0 Interesting
0 Brilliant
0 Confused

Compare the given expression to $\large (2^m)^2$ + $\large (2^p)^2$ + $\large 2(2^m)(2^p)$ = $\large (2^m + 2^p)^2$ , where $\large 2p = n$ . Two cases arise, one when $\large (2^{6012})$ is the $\large 2(2^m)(2^n)$ term and second when its a squared term.

You shall notice that the former case gives a larger value for n.

0 Helpful
0 Interesting
0 Brilliant
0 Confused

I think you didn't make clear why in the first case we get the largest possible $n$ .

Chris Galanis
- 5 years, 5 months ago

×

Problem Loading...

Note Loading...

Set Loading...