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Martin Sleziak
Martin Sleziak
5:16 AM
yesterday, by Martin Sleziak
I tried to mention this chat room sometimes in meta when some question or answer was related to the topic of this room. But it seems that did not attract too much attention.
As you have mentioned in your post, apart from reopen review queue there is also a thread on meta for reopen requests. (And there is also a dedicated chat room.) For the benefit of the users who are unaware of them I'll mention that links can be found, for example, in the re-open tag-info here on meta. — Martin Sleziak 54 secs ago
A message about this room I posted in the main chat room this week is starred, but perhaps after some time we can post another one.
That is, if we want to promote this room and get here more users.
 
 
6 hours later…
Martin Sleziak
Martin Sleziak
11:30 AM
duplicates I voted to close this questions as a duplicate - 457296: Prove that if $a$ and $b$ are relatively prime, then $\gcd(a+b, a-b) = 1$ or $2$ (timeline/revisions). In the review it got 3 votes to leave open and 3 votes to closed. Did I miss some reason why it should not be considered a duplicate?
Here is link to the review: math.stackexchange.com/review/close/847705
I am not sure whether it is needed, but I will also ping @DanielFischer since he interacted with the question. (In case you have something to say about that question. You were asked by the OP to post your comment as an answer.)
 
 
4 hours later…
user21820
user21820
3:56 PM
@MartinSleziak: Perhaps it's because those reviewers didn't read carefully enough? Though there's an easier solution: gcd(a+b,a-b) = gcd(a+b,2a) | gcd(2(a+b),2a) = gcd(2b,2a) = 2·gcd(b,a).
 
Martin Sleziak
Martin Sleziak
Possible, although I would be surprised to find out that Glorfindel is not a careful reviewer.
 
 
1 hour later…
Martin Sleziak
Martin Sleziak
5:22 PM
@user21820 BTW I think nobody will object if you add your solution as another answer.
 

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