I5.3 It is known that $b^{16}-1$ has four distinct prime factors, determine the largest one, denoted by $c$
$\begin{array}{l}2^{16}-1=(2-1)(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)=3 \times 5 \times 17 \times 257 \\ c=257\end{array}$