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Old 2016-05-01, 04:50   #1

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Default 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.
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Old 2016-05-01, 10:12   #2
R. Gerbicz
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"Robert Gerbicz"
Oct 2005

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 0<k<a (otherwise c<b) and gcd(k,a*b)=1, because k|(a*b-1).
a*c==1 mod b is also true, so
a*(a*b-1)/k==1 mod b
a*(-1)/k==1 mod b
a+k==0 mod b
hence b|k+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*b-1)/k=5.
(2,3,5) is really a solution, and there is no more.
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Old 2016-05-05, 06:42   #3
Romulan Interpreter
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Jun 2011

944710 Posts

Originally Posted by R. Gerbicz View Post
Nice puzzle. I'll give all solutions in (positive) integers.
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