Guest Posted March 23, 2010 Report Share Posted March 23, 2010 Each of A, B and C is a prime number with A <= B <= C. Determine all possible triplet(s) (A, B, C) such that: (I) A^{2} + B^{2} + C^{2} is a prime number, and: (II) C^{6} does not leave a remainder of 1 when divided by 14. Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted March 24, 2010 Report Share Posted March 24, 2010 Looking at condition (II), C^{6} mod 14 is always eight when even and always one when odd with the exception of odd multiples of 7 in which case it's seven. So, in order for it not to be one, C must be even or an odd multiple of seven. To qualify that C is also prime, it must be either 2 or 7. If C=2, in order for the sum of A^{2}+B^{2}+C^{2} to also be prime, it must be odd so exactly one of A or B must be odd and to satisfy A<=B<=C the only possibility for (A,B,C) where C=2 is (1,2,2) but then A^{2}+B^{2}+C^{2} = 9 which is not prime. So I think the only possibility for C is seven. From there all I gots is plug and chug. But by my count there are only 10 choices with C=7 and A<=B<=C and of those (3,3,7) (3,5,7) and (3,7,7) are the only (A,B,C) that satisfy the conditions of the OP. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted March 24, 2010 Report Share Posted March 24, 2010 (edited) Each of A, B and C is a prime number with A <= B <= C. Determine all possible triplet(s) (A, B, C) such that: (I) A^{2} + B^{2} + C^{2} is a prime number, and: (II) C^{6} does not leave a remainder of 1 when divided by 14. Test k^6 mod 14 for k in 0..13 to get the sequence 0,1,8,1,8,1,8,7,8,1,8,1,8,1 The C's of interest will therefore with respect to modulus 14 be congruent to 0,2,4,6,7,8,10,12 This means C can be any even number or number of the form 7*(2n+1) for integer n. The only prime even number is 2. The only prime number of the form 7*(2n+1) is 7. C must be either 2 or 7. If C is 2, A and B, being prime numbers must also be 2. However, the sum of these squares gives 12 which is not prime, so C cannot be 2. C must be 7 then. A and B make take values of primes below and including 7 which are 2,3,5,7. Let us consider the cases and the sum of the squares: A,B,SumofSquares,IsPrime 2,2,57,No 2,3,62,No 2,5,78, No 2,7,102,No 3,3,67,Yes 3,5,83,Yes 3,7,107,Yes 5,5,99,No 5,7,123,No 7,7,147,No Thus the triplets are 3,3,7 3,5,7 3,7,7 Checking plainglazed's answer, I see that we agree. Edited March 24, 2010 by mmiguel1 Quote Link to comment Share on other sites More sharing options...

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Each of A, B and C is a prime number with A <= B <= C.

Determine all possible triplet(s) (A, B, C) such that:

(I) A

^{2}+ B^{2}+ C^{2}is a prime number, and:(II) C

^{6}does not leave a remainder of 1 when divided by 14.## Link to comment

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