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Sum (Squares) = Prime Number (Posted on 2010-09-09) Difficulty: 2 of 5
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.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (2 votes)

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analytical solution | Comment 1 of 2

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


  Posted by Daniel on 2010-09-09 13:09:42
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