2^1024 -1 = 3 x 5 x 17 x 257 x 641 x 65537 x 274177 x 2424833 x 6700417 x 67280421310721.........etc
+26367 What are the two smallest prime factors of \(2^{1024} - 1\)?
\(\begin{array}{|rcll|} \hline && 2^{1024} - 1 \\ &=& \left( 2^{512} - 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{256} - 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{128} - 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{64} - 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{32} - 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{16} - 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{8} - 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{4} - 1\right) \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{2} - 1\right) \left( 2^{2} + 1\right) \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{1} - 1\right) \left( 2^{1} + 1\right) \left( 2^{2} + 1\right) \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& 1\times 3 \times 5 \times \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ \hline \end{array} \)
The two smallest prime factors are 3 and 5, they are also the smallest possible odd prime numbers,
all factors are odd,
so that the 2 as the only even prime number cannot be included
