Exercise 3. Let $a$ and $b$ be integers such that $0 < b < a$ and

$$
\frac{2^{8060}-1}{\left(2^{4030}+1\right)\left(2^{2015}-1\right)}=2^{a}+b
$$

Determine the value of $a+b$.
Solution of the exercise $32^{8060}-1=\left(2^{4030}\right)^{2}-1=\left(2^{4030}-1\right)\left(2^{4030}+1\right)=\left(2^{2015}-1\right)\left(2^{2015}+\right.$ $1)\left(2^{4030}+1\right)$, so the fraction is equal to $2^{2015}+1$. Since $2^{a}<2^{a}+b<2^{a}+a<2^{a+1}$, we deduce $2^{a}<2^{2015}+1<2^{a+1}$, thus $a \leqslant 2015<a+1$. The only possibility is $a=2015$ and $b=1$, so $a+b=2016$.