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.
I first started with condition ii
The possible values for x^6 mod 14 is 0,1,8,7
for values of x<14 any remainder other than 1 implies that x is not prime as it would share a common factor with 14. Now the only primes less than 14 which do not give x^6 mod 14=1 are 2 and 7, now if c=2 then a=b=c=2 and thus a^2+b^2+c^2 is not prime. Thus c=7
Now since we can not have a=b=c then this gives the following possible solutions
(2,2,7), (2,3,7), (2,5,7), (2,7,7)
(3,3,7), (3,5,7), (3,7,7)
(5,5,7), (5,7,7)
of which only
(3,3,7), (3,5,7), and (3,7,7) satisfy condition i
for which the prime number attained is
3^2+3^2+7^2=67
3^2+5^2+7^2=83
3^2+7^2+7^2=107
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Posted by Daniel
on 2010-09-09 13:09:42 |
analytical solution
