What is the sum of all distinct prime factors of
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We make use of the following identities:
To start, we rewrite the sum of the three center terms as 4 7 + 5 5 + 6 4 = ( 5 4 + 6 4 ⋅ 4 4 ) + ( 6 4 + 4 ⋅ 5 4 ) . From here we apply Identity 2 to the first group and Identity 1 to the second to obtain the following factorizations. 5 4 + 6 4 ⋅ 4 4 6 4 + 4 ⋅ 5 4 = ( 5 2 + 4 ⋅ 5 ⋅ 4 + 8 ⋅ 4 2 ) ( 5 2 − 4 ⋅ 5 ⋅ 4 + 8 ⋅ 4 2 ) = ( 2 3 3 ) ( 7 3 ) = ( 6 2 + 2 ⋅ 6 ⋅ 5 + 2 ⋅ 5 2 ) ( 6 2 − 2 ⋅ 6 ⋅ 5 + 2 ⋅ 5 2 ) = ( 1 4 6 ) ( 2 6 ) As for the remaining two terms, we can apply Identity 3: 3 3 + 7 6 = ( 3 + 7 2 ) ( 3 2 − 3 ⋅ 7 2 + 7 4 ) = ( 5 2 ) ( 2 2 6 3 ) = ( 5 2 ) ( 3 1 ⋅ 7 3 ) We see that 7 3 is a common factor of all three of these sums. Factoring it out, we have 3 3 + 4 7 + 5 5 + 6 4 + 7 6 = 7 3 ( 2 3 3 + 5 2 ( 1 + 3 1 ) ) = 7 3 ( 1 8 9 7 ) . Since 7 3 is prime, it remains to factor 1 8 9 7 ; trying small factors yields the prime factorization 1 8 9 7 = 7 ⋅ 2 7 1 . This gives 7 + 7 3 + 2 7 1 = 3 5 1 to be the answer.