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# 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

Note by Rakhi Bhattacharyya
4 months, 2 weeks ago

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This is IMO 1990 P3. See here. · 4 months, 1 week ago

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.) · 4 months, 2 weeks ago

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

1. 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 Staff · 3 months, 3 weeks ago

How do you know $$a$$ is an integer? · 4 months, 1 week ago

n=1 is also an answer. Put k=9. U believe these are the only 2 solutions. · 4 months, 2 weeks ago

$$k=3$$ and $$n=1$$ also works. · 4 months, 2 weeks ago