[abetusk]

I'm wrong. The Mahler-Mignotte only works for pairs of roots and doesn't say anything about the absolute value of the sum, at least in the way I was thinking about it. There may be a way to "fix it up" but not that I see and I suspect folks who've studied this in earnest are aware of Mahler-Mignotte and understand why it can't be used to solve this problem.

Thanks to @devit [0] who has understood why the Mahler-Mignotte tactic doesn't work. Just because you can bound the bit complexity of pairs of roots doesn't mean you can bound the complexity of all the 2^N possible {-1,1} combinations of them. At least, I don't see how it can be done simply.

[0] https://news.ycombinator.com/item?id=30059545