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We denote by $p(n)$ the number of partitions of $n$. There are infinitely many integers $m$ such that $p(m)$ is even, and infinitely many integers $n$ such that $p(n)$ is odd.

It might be proved by the Euler's Pentagonal Number Theorem. Could you give me some hints?

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    I can imagine various ways to interpret "the number of partitions of $n$". Which one is in use here?2011-10-27
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    Just a quickly hint, need to go now. Try search about Ramanujan too and try this [link](http://en.wikipedia.org/wiki/Partition_(number_theory))2011-10-27
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    @Henning: Both the notation and the reference to Euler’s pentagonal number theorem strongly suggest that it’s the usual notion of [partition of an integer](http://en.wikipedia.org/wiki/Partition_%28number_theory%29).2011-10-27

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Hint: Look at the pentagonal number theorem and the recurrence relation it yields for $p(n)$, and consider what would happen if $p(n)$ had the same parity for all $n\gt n_0$.

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The theorem is due to O. Kolberg, Note on the parity of the partition function, Math. Scand. 7 1959 377–378, MR0117213 (22 #7995). The review by Mirsky says that the proof uses the Pentagonal Number Theorem, and is extremely short and simple.

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    I'm surprised it's considered to be due to anyone. It follows almost immediately from the pentagonal number theorem.2011-10-27
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    @joriki, the reviewer was surprised, too, that the theorem had been overlooked by earlier writers.2011-10-27