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## Question

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.

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Looking at condition (II), C6 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 A2+B2+C2 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 A2+B2+C2 = 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.

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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) A2 + B2 + C2 is a prime number, and:

(II) C6 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 by mmiguel1

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