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Dec 31, 2010 at 9:56 comment added Wadim Zudilin Luis, you are kidding! There was already a post which has recognized 2 to be the oddest integer. :-) Too many constraints. Please try yourself to modify everything for $4\cdot3^2$.
Dec 31, 2010 at 7:54 comment added Luis H Gallardo $16 = 4 \cdot 2^2$ i.e., $h=2$ that is not odd
Dec 31, 2010 at 3:51 comment added Wadim Zudilin I finally realized that there was a pre-question to this question already answered by the (same) unknown. And I would join the above remark of the unknown: it is a strange question without sufficient motivation...
Dec 31, 2010 at 3:42 history edited Wadim Zudilin CC BY-SA 2.5
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Dec 31, 2010 at 2:38 comment added Igor Rivin I believe the question asks for $\pm 1$ coefficients, so this is not a counterexample.
Dec 31, 2010 at 2:28 history answered Wadim Zudilin CC BY-SA 2.5