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Find all integers $n$ such that $\dfrac{2^n+1}{n^2}$ is also an integer.

(IMO-$1990$)

Inceptio
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Carlos Beris
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    n is not even because even number cant divide odd number – M.H Jun 29 '13 at 05:47
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    http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln903.html – lab bhattacharjee Jun 29 '13 at 06:03
  • They're asking for integers, not primes. $171$ divides $2^{171}+1$, for example. The number $n$ can be a product of primes in $1$ and $3$ modulo $8$, as would its period. The answers that i have seen in the url above and the previous question consider only primes, not composite integers. – wendy.krieger Jun 29 '13 at 08:59
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    @wendy.krieger But $171^2 \not\mid 2^{171}+1$, which is what the question is asking about. The solution posted to the duplicate is indeed correct for all $n$. It uses a prime factor of $n$, which is guaranteed to exist. – Alex Becker Jun 29 '13 at 18:16
  • @AlexBecker But it only handles base $2$, and some of the versions are seriously flawed and all are overly complex. A general proof and construction is given for all $b$, including $10$. – wendy.krieger Jun 30 '13 at 08:03

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