Guess the prime factors - 2

The given number is slightly less than ${10}^{12}$ and is in fact ${10}^{12}-4096={10}^{12}-{2}^{12}$ .

This factors as ${2}^{12}({5}^{12}-1)$ and ${5}^{12}-1=({5}^{3}+1)({5}^{3}-1)({5}^{2}+1)({5}^{4}-{5}^{2}+1)$

$=(5-1)({5}^{2}+5+1)(5+1)({5}^{2}-5+1)({5}^{2}+1)({5}^{4}-{5}^{2}+1)$

$={ 2 }^{ 2 }\cdot 31\cdot 2\cdot 3\cdot 3\cdot 7\cdot 2\cdot 13\cdot 601$

Thus, the given number is ${ 2 }^{ 16 }\cdot { 3 }^{ 2 }\cdot 7\cdot 13\cdot 31\cdot 601$ and its largest prime factor is $\boxed{601}$ .