mersenneforum.org Another "interesting" puzzle
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 2016-05-01, 04:50 #1 Trejack   23×251 Posts Another "interesting" puzzle I've come up with another "hard" puzzle I would like to share (and this is my second problem to come up with): Are there a set of three primes a, b, and c such that: (a*b) = 1 (mod c) (a*c) = 1 (mod b) (b*c) = 1 (mod a) besides 2, 3, 5. I tried testing primes b, c with a = 2, and I could not complete the equation. Anyone else have an idea to this? Thanks.
 2016-05-01, 10:12 #2 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 2·733 Posts Nice puzzle. I'll give all solutions in (positive) integers. We can assume that a<=b<=c. If a=1 then it is easy to see that b=1 and c can be anything. So (a=b=1,c) is a solution and its permutations. Now, we can say that a>1, then from a*b==1 mod c we get that c=(a*b-1)/k, where 01, so a=2, from this b=k+a, but 0
2016-05-05, 06:42   #3
LaurV
Romulan Interpreter

Jun 2011
Thailand

944710 Posts

Quote:
 Originally Posted by R. Gerbicz Nice puzzle. I'll give all solutions in (positive) integers.
Beautiful!

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