Question 206, Given the pair of positive integers $(a, b)$ with the greatest common divisor $(a, b)=2015$, find the minimum value of $a+b$ such that $(a+b) \mid\left[(a-b)^{2016}+b^{2016}\right]$.
Question 206, Solution: Let $a=2015c, b=2015d$, then $(c, d)=1$. Thus, $a+b=2015(c+d)$, and
$(c+d) \mid \left[(c-d)^{2016}+d^{2016}\right] \cdot 2015^{2015}$
Since $(c-d)^{2016}+d^{2016} \equiv (-2d)^{2016}+d^{2016} \equiv (2^{2016}+1)d^{2016} \pmod{c+d}$, it follows that $(c+d) \mid (2^{2016}+1)d^{2016} \cdot 2015^{2015}$
$(c+d) \mid (2^{2016}+1) \cdot 2015^{2015}$
Noting that the smallest true divisor of $(2^{2016}+1) \cdot 2015^{2015}$ is 5, we have $c+d \geq 5$. Therefore, $a+b=2015(c+d) \geq 2015 \times 5 = 10075$.
On the other hand, when $a=2015, b=8060$, we have $a+b=10075$.
In conclusion, the minimum value of $a+b$ that satisfies the condition is 10075.