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) A2 + B2 + C2 is a prime number, and:
(ii) C6 does not leave a remainder of 1 when divided by 14.
As Daniel rightly points out, condition 2 means in substance that c is 7 or a number oddly divisible by 7 (so not prime). C is therefore 7. All 3 squares cannot be the same, as P would then be divisible by 3. Further, P cannot be of the form 4k+1, since this would require a, b to be even. a, b would then have to be 2, giving the result 57, which is not prime. Since primes that are sums of 3 squares are congruent to 1,3,or 5 mod 8, P must be of the form 8k+3. This gives:
(2k1-1)^2+(2k2-1)^2=8k-46 k1>1,k2>1, k<18
for which there are 11 integer solutions, of which all but 3,3,7,67; 3,5,7,83; and 3,7,7,107 can readily be eliminated.
Edited on September 10, 2010, 12:21 am
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Posted by broll
on 2010-09-09 16:25:01 |
solution
