Be observant!

By Sophie-Germain identity,

$a^4+4b^4 = (a^2+2ab+2b^2)(a^2-2ab+2b^2)$

$629 = 625+4 = 5^4+4 \cdot 1^4$

This can be factored out to $(5^2+2 \cdot 5 +2 \cdot 1^2)(5^2-2 \cdot 5 +2 \cdot 1^2)$

This is equal to $17 \cdot 37$ . Since $17$ and $37$ are prime, the sum of all the distinct prime factors of 629 is $17+37 = 54$

Prime factorization of $629$ is $17\times37$ therefore answer must be $17+37=\boxed{54}$