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Nov 30, 2016 at 16:31 vote accept T. Amdeberhan
Nov 30, 2016 at 16:13 comment added GH from MO @T.Amdeberhan: Specifying an explicit lower bound for $n$ seems doable, but it will involve considerable amount of work. The main task is to calculate explicitly the implied constants in (349) and (352) of Heath-Brown & Tolev, and for that (I think) one needs to study a big portion of the paper in detail. If you are that much interested in this problem, use it as motivation to study Heath-Brown & Tolev.
Nov 30, 2016 at 12:35 comment added T. Amdeberhan @GHfromMO: very nice. I wonder how specific can "large $n$" be specified so that the rest might be checked by computer. This would make a complete resolution.
Nov 29, 2016 at 23:23 comment added GH from MO @FedorPetrov: I think you are right. I focused on the actual proportions in my post, and it is for the $o(1)$ terms that we need a sufficiently large odd part of $N$. But in the end, as you say, the $2$-part of $N$ is irrelevant for the odd gcd of the products $a_1a_2a_3a_4$.
Nov 29, 2016 at 22:31 comment added Fedor Petrov If it works for large enough odd numbers, it works for all numbers with large enough odd part by some obvious reasons (multiplying by 2 works as $2(a^2+b^2+c^2+d^2)=(a+b)^2+(a-b)^2+(c+d)^2+(c-d)^2$, and the greatest odd common divisor of $a+b,a-b,c+d,c-d$ is the same as for $a,b,c,d$).
Nov 29, 2016 at 21:35 comment added GH from MO @FedorPetrov: I think this works for arbitrary numbers $N$ (instead of $N=4n+1$) as long as the odd part of $N$ exceeds $N^c$ for some fixed $c>3/4$.
Nov 29, 2016 at 20:57 comment added Fedor Petrov Does this work for numbers $4n+3$ instead of $4n+1$?
Nov 29, 2016 at 18:17 history edited GH from MO CC BY-SA 3.0
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Nov 29, 2016 at 17:55 history edited GH from MO CC BY-SA 3.0
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Nov 29, 2016 at 17:49 history answered GH from MO CC BY-SA 3.0