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# help

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What are the two smallest prime factors of 2^{1024} - 1?

Nov 15, 2019

#1
+2850
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Hint: $$(a^2-b^2)=(a+b)(a-b)$$

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Nov 15, 2019
#2
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2^1024 -1 = 3 x 5 x 17 x 257 x 641 x 65537 x 274177 x 2424833 x 6700417 x 67280421310721.........etc

Nov 15, 2019
#3
+24430
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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

Nov 15, 2019