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 12h accepted Tangent bundle of manifold with no odd dimensional sub-bundles 12h comment Tangent bundle of manifold with no odd dimensional sub-bundles I realized after asking the first question that it was rather obviously false; thanks for the simple counterexample. I guess that the orientable double cover is an odd-dimensional orientable subbundle of the double cover of $M$, which is impossible since it's entirely contained in one of the copies of $M$. 2d asked Tangent bundle of manifold with no odd dimensional sub-bundles 2d accepted $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes 2d comment $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes Thanks! I see the mistake I was making before - I didn't check that $(1+ax+x^2)^{-1}(1+x)^{n+1}$ could have degree less than $n-1$ in larger cases. Quite silly of me 2d revised $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes added info 2d asked $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes Mar23 awarded Popular Question Mar4 comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$ that being said, now it's unclear what the problem statement actually means, though we assume the more difficult one evidently Mar4 revised Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$ deleted 15 characters in body Mar4 comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$ @rah4927, I think you're right; my original line of thinking was that if they're all distinct, then the LHS is bounded by something smaller than the RHS, but I think I messed up. I'll revert it. Mar3 comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$ yes, you're right. my mistake! Mar3 comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$ For $n=3,$ there are two solutions: $(1,2,3)$ and $(2,2,2)$ so wolframalpha lied to you. @rah4927, yes, there are infinitely many solutions over the reals. Just pick $a_i \le i$ for $i < n$ and then by continuity you can always find $a_n$ that works. Mar3 comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$ A remark: $a_i\mid 2i$ for each $i$ and note that solutions with $a_1 = 1$ are in bijective correspondence with solutions having $a_1 = 2$ just by taking the $n$-tuple with $2i/a_i$ in place of $a_i,$ so you can assume $a_1 = 1$ without loss of generality. Also, you can reduce it to an integer equation, but I don't know how helpful that is. Mar3 revised Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$ clarifying problem statement, as $a_i$'s are not necessarily distinct Mar2 answered Minimizing the product $xy$ subject to a polynomial constraint on $x, y$ Mar1 reviewed Approve Complex Number Maths Feb14 comment Bound on $L^1$ norm of pairwise sums yes @RoryDaulton Feb14 revised Bound on $L^1$ norm of pairwise sums expanded on question Feb14 comment Bound on $L^1$ norm of pairwise sums haha yes, sorry