20160501, 04:50  #1 
11550_{8} 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. 
20160501, 10:12  #2 
"Robert Gerbicz"
Oct 2005
Hungary
11000100111_{2} 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*b1)/k, where 0<k<a (otherwise c<b) and gcd(k,a*b)=1, because k(a*b1). a*c==1 mod b is also true, so a*(a*b1)/k==1 mod b a*(1)/k==1 mod b a+k==0 mod b hence bk+a, but 0<k+a<2*a<=2*b, so only b=k+a is possible. Use the last equation: b*c==1 mod a (k+a)*(a*(k+a)1)/k==1 mod a k*(1)/k==1 mod a 2==0 mod a, but a>1, so a=2, from this b=k+a, but 0<k<2, so k=1 and b=k+a=3, c=(a*b1)/k=5. (2,3,5) is really a solution, and there is no more. 
20160505, 06:42  #3 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10,091 Posts 

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