A good problem

This is a problem I recently found which I could not solve. This was there in a book prescribed for RMO and INMO. Here it is:

Determine with proof the set of all positive integers $n$ such that $2^n +1$ is divisible by $n^2$ .

Any hint, or solutions are welcome

#NumberTheory

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Comments

Since, $2^{n} +1$ is divisible by $n^2$ .

Let,
$2^{n}+1= kn^2$
$\implies 2^n = (\sqrt{k}n+1)(\sqrt{k}n-1)$
Let, $\sqrt{k}n-1 = a$
Then,
$2^n = (a)(a+2)$
Now,
We have $2$ powers of $2$ with a difference of $2$ ;that add up to an integer value;
Which, must be $2$ and $4$
Hence, $n=3$ and $k=1$ is one answer.

Moreover, as @Kushagra Sahni has pointed out; $n=1$ also works. I cant find a rigorous proof approach to $n=1$ so, for now I will be satisfied by the fact that since 1 divides every natural number it divides $2^n +1$ .

P,S.: I am not very good at these kind of proofs so, there may be a trick or two I have missed and maybe even another value for $n$ (though, I somehow feel quite sure about this answer.)

Yatin Khanna - 4 years, 9 months ago

n=1 is also an answer. Put k=9. U believe these are the only 2 solutions.

Kushagra Sahni - 4 years, 9 months ago

Told you I tend to miss few tricks...
$k=3$ and $n=1$ also works.

Yatin Khanna - 4 years, 9 months ago

How do you know $a$ is an integer?

Ameya Daigavane - 4 years, 9 months ago

Thanks for putting in the effort to write up the solution. Unfortunately, it is incorrect because of:

You are making the assumption that $k$ must be a perfect square, so that $\sqrt{k} n - 1 = a$ is an integer.

Keep up the good effort! Practice makes perfect

Calvin Lin Staff - 4 years, 8 months ago

This is IMO 1990 P3. See here.

Ameya Daigavane - 4 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$