Cyclic congruent
How many triples of positive integers ( a , b , c ) satisfy a , b , c > 1 and the following?
⎩ ⎨ ⎧ a + b ≡ 1 ( m o d c ) b + c ≡ 1 ( m o d a ) c + a ≡ 1 ( m o d b )
If you think that there are infinitely many such triples, submit -1 as your answer.
The answer is 15.
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1 solution
Why b+c-1 <2a ?
Umm.. I have a bogus solution and need some clarification as to what is wrong...please help:
We convert each of the given congruences into linear equation as shown:
a + b = c x + 1 , b + c = a y + 1 & b z + 1 = c + a . Then, we manipulate this linear system of equations to obtain c x − a y = a − c , a y − b z = b − a & c x − b z = b − c . An obvious solution for ( x , y , z ) would be ( − 1 , − 1 , − 1 ) . We then substitute these values into our initial equations and get the result a + b = 1 − c , b + c = 1 − a & c + a = 1 − b . Adding these up gives 3 ( a + b + c ) = 3 implying a + b + c = 1 . But this cannot be true as ( a , b , c ) > 1 are positive integers.
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a , b , c > 1 hence x , y , z > 0 .
So, x = y = z = − 1 is invalid.
WLOG, assume that a ≥ b ≥ c > 1 .
We have 0 < b + c − 1 < 2 a and a ∣ b + c − 1 , so b + c − 1 = a .
Hence, b ∣ ( b + c − 1 ) + c − 1 or b ∣ 2 c − 2 . Since 0 < 2 c − 2 < 2 b , we get b = 2 c − 2 , and a = 3 c − 3 .
Since c ∣ a + b − 1 or c ∣ 5 c − 6 , we get c ∣ 6 , implies that c ∈ { 2 , 3 , 6 } .
If c = 2 then a = 3 , b = 2 , there are 3 permutations in this case.
If c = 3 then a = 6 , b = 4 , there are 6 permutations in this case.
If c = 6 then a = 1 5 , b = 1 0 , there are 6 permutations in this case.
So there are 1 5 such triples.