Factoring a Sum of Five Perfect Powers?

We make use of the following identities:

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

To start, we rewrite the sum of the three center terms as $4^7 + 5^5 + 6^4 = (5^4 + 64 \cdot 4^4) + (6^4 + 4 \cdot 5^4)$ . From here we apply Identity 2 to the first group and Identity 1 to the second to obtain the following factorizations. $\begin{aligned} 5^4 + 64 \cdot 4^4 & = (5^2 + 4 \cdot 5 \cdot 4 + 8 \cdot 4^2)(5^2 - 4 \cdot 5 \cdot 4 + 8 \cdot 4^2) \\ & = (233)(73) \\6^4 + 4 \cdot 5^4 & = (6^2 + 2\cdot 6 \cdot 5 + 2 \cdot 5^2)(6^2 - 2\cdot 6 \cdot 5 + 2 \cdot 5^2) \\& = (146)(26) \end{aligned}$ As for the remaining two terms, we can apply Identity 3: $\begin{aligned} 3^3 + 7^6 & = (3 + 7^2)(3^2 - 3 \cdot 7^2 + 7^4) \\ & = (52)(2263) \\ & = (52)(31 \cdot 73) \end{aligned}$ We see that $73$ is a common factor of all three of these sums. Factoring it out, we have $3^3 + 4^7 + 5^5 + 6^4 + 7^6 = 73(233 + 52(1 + 31)) = 73(1897)$ . Since $73$ is prime, it remains to factor $1897$ ; trying small factors yields the prime factorization $1897 = 7 \cdot 271$ . This gives $7 + 73 + 271 = \boxed{351}$ to be the answer.