For a subset , let . Show that there is auniquepartition of such that and contain no odd prime.

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- December 4th 2009, 04:13 PMBruno J.A nice problem for fun - Iranian math competition
For a subset , let . Show that there is a

**unique**partition of such that and contain no odd prime. - December 4th 2009, 04:22 PMlvleph
So I am thinking this has something to do with being an ideal in . And that every ideal of is of the form where . I believe we may want . That is my thoughts right off hand.

- December 4th 2009, 04:53 PMJose27
Let denote the partition of in even and odd integers (resp), and let be any other partition that satisfies the condition then (say) but then clearly because odd+even=odd and so since cannot be contained in , and by the same argument and since they form a partition and .

- December 4th 2009, 06:09 PMBruno J.
Why "clearly" ? Odd numbers are allowed, just not odd primes. Nowhere do you use the prime-related condition. You just replaced it by a stronger condition, which, as you noticed, makes the problem quite trivial.

- December 4th 2009, 09:56 PMDrexel28
- December 5th 2009, 05:27 AMNonCommAlg
i wouldn't call Bertrand's postulate a big-gun theorem because the proof (Erdos) of this theorem is fairly short and very elementary. actually, students who participate in math competetions

(even IMO, never mind Putnam!) are expected to know and use it. anyway, i just wanted to add this to Drexel28's comment that the problem is easily solved using Bertrand's postulate. - December 5th 2009, 05:30 AMDrexel28
- December 6th 2009, 08:33 PMBruno J.
My solution makes use of Bertrand's postulate. Perhaps there is another way. Good job!

- December 7th 2009, 10:06 PMDrexel28