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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? –  Henning Makholm Oct 27 '11 at 1:08
    
Just a quickly hint, need to go now. Try search about Ramanujan too and try this link –  GarouDan Oct 27 '11 at 1:11
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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. –  Brian M. Scott Oct 27 '11 at 4:54

2 Answers 2

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. –  joriki Oct 27 '11 at 4:59
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@joriki, the reviewer was surprised, too, that the theorem had been overlooked by earlier writers. –  Gerry Myerson Oct 27 '11 at 5:28

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