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$\begin{aligned} (a+b)^7 - a^7 - b^7 & \equiv 0 \text{ (mod 7}^7) \\ 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 & \equiv 0 \text{ (mod 7}^\blue 7) \\ a^6b + 3a^5b^2 + 5a^4b^3 + 5a^3b^4 + 3a^2b^5 + ab^6 & \equiv 0 \text{ (mod 7}^\red 6) & \small \blue{\text{Since }ab\text{ is indivisible by 7}} \\ a^5 + 3a^4b + 5a^3b^2 + 5a^2b^3 + 3ab^4 + b^5 & \equiv 0 \text{ (mod 7}^6) \\ a^5 + b^5 + 3ab(a^3+b^3) + 5a^2b^2(a+b) & \equiv 0 \text{ (mod 7}^6) \\ (a + b)(a^4-a^3b+a^2b^2-ab^3+b^4) + 3ab(a+b)(a^2-ab+b^2) + 5a^2b^2(a+b) & \equiv 0 \text{ (mod 7}^6) & \small \blue{\text{Since }a+b\text{ is indivisible by 7}} \\ a^4-a^3b+a^2b^2-ab^3+b^4 + 3ab(a^2-ab+b^2) + 5a^2b^2 & \equiv 0 \text{ (mod 7}^6) \\ a^4 +b^4 + 2ab(a^2+b^2) + 3a^2b^2 & \equiv 0 \text{ (mod 7}^6) \\ (a^2 +b^2)^2 + 2ab(a^2+b^2) + a^2b^2 & \equiv 0 \text{ (mod 7}^6) \\ \left(a^2 +b^2 + ab\right)^2 & \equiv 0 \text{ (mod 7}^\blue 6) \\ a^2 +b^2 + ab & \equiv 0 \text{ (mod 7}^\red 3) \end{aligned}$

Assuming that

$\begin{aligned} a^2 +b^2 + ab & = 7^3 = 343 \\ \implies b^2 + ab + a^2 - 343 & = 0 \\ \implies b & = \frac {-a + \sqrt{1372-3a^2}}2 \end{aligned}$

For $b$ to be an integer, $1372-3a^2$ must be a perfect square. Solving for $a$ , we have the unordered pairs of $(a,b) = (-7, 21), (-1, 19), \boxed{(1,18)}, (7,14)$ . Therefore there is only one acceptable solution and $a+b = 1 + 18 = \boxed{19}$ .