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Jul 11, 2020 at 21:16 comment added Per Alexandersson They use $d_m$ as the GCD of $a_1,\dotsc,a_m$. One can then quite easily factor out this from the problem, and assume $a_1=1$ without losing much.
Jul 11, 2020 at 20:56 history edited Carl-Fredrik Nyberg Brodda CC BY-SA 4.0
Removed some examples after helpful comment.
Jul 11, 2020 at 19:54 comment added Carl-Fredrik Nyberg Brodda @MaxAlekseyev You are right, of course. I misread his condition as pairwise co-prime. Nevertheless, the main point is that no additional assumptions are added inside the text to the statement of the theorem.
Jul 11, 2020 at 12:20 comment added Max Alekseyev I'm confused then. Your bullet points do not support this conclusion (only that they assume no pairwise coprimality), and according to @Ramanumpy, all their examples contain 1 and thus are co-prime.
Jul 11, 2020 at 7:43 comment added Carl-Fredrik Nyberg Brodda @MaxAlekseyev No, I mean it as it is written; "their many examples clearly show that they do not assume [coprimality of the $a_i$]".
Jul 11, 2020 at 2:16 comment added Max Alekseyev "their many examples clearly show that they do not assume this" - do you mean they do assume this by any chance?
Jul 10, 2020 at 20:10 comment added Ramanumpy Unfortunately it was precisely the examples that made me suspicious because they are all coprime cases due to the first value 1 (common to all the examples). The lack of explicit declaration of coprimality and the observation you make on page 57 I would say that they close the discussion making an even more remarkable a beautiful theorem. Thanks.
Jul 10, 2020 at 20:00 vote accept Ramanumpy
Jul 10, 2020 at 19:40 history answered Carl-Fredrik Nyberg Brodda CC BY-SA 4.0