Factoring a Sum of Five Perfect Powers?

What is the sum of all distinct prime factors of 3 3 + 4 7 + 5 5 + 6 4 + 7 6 ? 3^3 + 4^7 + 5^5 + 6^4 + 7^6?


The answer is 351.

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1 solution

Brian Yao
Aug 30, 2018

We make use of the following identities:

  • Identity 1 (The Sophie Germain Identity): a 4 + 4 b 4 = ( a 2 + 2 a b + 2 b 2 ) ( a 2 2 a b + 2 b 2 ) a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)
  • Identity 2 (Variant of the SGI): a 4 + 64 b 4 = ( a 2 + 4 a b + 8 b 2 ) ( a 2 4 a b + 8 b 2 ) a^4 + 64b^4 = (a^2 + 4ab + 8b^2)(a^2 - 4ab + 8b^2)
  • Identity 3 (Sum of cubes): a 3 + b 3 = ( a + b ) ( a 2 2 a b + b 2 ) a^3 + b^3 = (a + b)(a^2 - 2ab + b^2)

To start, we rewrite the sum of the three center terms as 4 7 + 5 5 + 6 4 = ( 5 4 + 64 4 4 ) + ( 6 4 + 4 5 4 ) 4^7 + 5^5 + 6^4 = (5^4 + 64 \cdot 4^4) + (6^4 + 4 \cdot 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 + 64 4 4 = ( 5 2 + 4 5 4 + 8 4 2 ) ( 5 2 4 5 4 + 8 4 2 ) = ( 233 ) ( 73 ) 6 4 + 4 5 4 = ( 6 2 + 2 6 5 + 2 5 2 ) ( 6 2 2 6 5 + 2 5 2 ) = ( 146 ) ( 26 ) \begin{aligned} 5^4 + 64 \cdot 4^4 & = (5^2 + 4 \cdot 5 \cdot 4 + 8 \cdot 4^2)(5^2 - 4 \cdot 5 \cdot 4 + 8 \cdot 4^2) \\ & = (233)(73) \\6^4 + 4 \cdot 5^4 & = (6^2 + 2\cdot 6 \cdot 5 + 2 \cdot 5^2)(6^2 - 2\cdot 6 \cdot 5 + 2 \cdot 5^2) \\& = (146)(26) \end{aligned} 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 ) = ( 52 ) ( 2263 ) = ( 52 ) ( 31 73 ) \begin{aligned} 3^3 + 7^6 & = (3 + 7^2)(3^2 - 3 \cdot 7^2 + 7^4) \\ & = (52)(2263) \\ & = (52)(31 \cdot 73) \end{aligned} We see that 73 73 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 = 73 ( 233 + 52 ( 1 + 31 ) ) = 73 ( 1897 ) 3^3 + 4^7 + 5^5 + 6^4 + 7^6 = 73(233 + 52(1 + 31)) = 73(1897) . Since 73 73 is prime, it remains to factor 1897 1897 ; trying small factors yields the prime factorization 1897 = 7 271 1897 = 7 \cdot 271 . This gives 7 + 73 + 271 = 351 7 + 73 + 271 = \boxed{351} to be the answer.

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