Prime Factors of 2^64 - 1
Jump to navigation
Jump to search
Theorem
The prime decomposition of $2^{64} - 1$ is given by:
- $2^{64} - 1 = 3 \times 5 \times 17 \times 257 \times 641 \times 65 \, 537 \times 6 \, 700 \, 417$
Proof
\(\ds 2^{64} - 1\) | \(=\) | \(\ds \paren {2^{32} }^2 - 1\) | Power of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {2^{32} - 1}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {\paren {2^{16} }^2 - 1}\) | Power of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {2^{16} + 1} \paren {2^{16} - 1}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {2^{16} + 1} \paren {\paren {2^8}^2 - 1}\) | Power of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {2^{16} + 1} \paren {2^8 + 1} \paren {2^8 - 1}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {2^{16} + 1} \paren {2^8 + 1} \paren {\paren {2^4}^2 - 1}\) | Power of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {2^{16} + 1} \paren {2^8 + 1} \paren {2^4 + 1} \paren {2^4 - 1}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {2^{16} + 1} \paren {2^8 + 1} \paren {2^4 + 1} \paren {\paren {2^2}^2 - 1}\) | Power of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {2^{16} + 1} \paren {2^8 + 1} \paren {2^4 + 1} \paren {2^2 + 1} \paren {2^2 - 1}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{32} + 1} \paren {2^{16} + 1} \paren {2^8 + 1} \paren {2^4 + 1} \paren {2^2 + 1} \paren {2 + 1} \paren {2 - 1}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^{\paren {2^5} } + 1} \paren {2^{\paren {2^4} } + 1} \paren {2^{\paren {2^3} } + 1} \paren {2^{\paren {2^2} } + 1} \paren {2^{\paren {2^1} } + 1} \paren {2^{\paren {2^0} } + 1}\) |
From Sequence of Fermat Primes:
\(\ds 2^{\paren {2^0} } + 1\) | \(=\) | \(\ds 3\) | ||||||||||||
\(\ds 2^{\paren {2^1} } + 1\) | \(=\) | \(\ds 5\) | ||||||||||||
\(\ds 2^{\paren {2^2} } + 1\) | \(=\) | \(\ds 17\) | ||||||||||||
\(\ds 2^{\paren {2^3} } + 1\) | \(=\) | \(\ds 257\) | ||||||||||||
\(\ds 2^{\paren {2^4} } + 1\) | \(=\) | \(\ds 65 \, 537\) |
all of which are Fermat primes.
Then:
\(\ds 2^{\paren {2^5} } + 1\) | \(=\) | \(\ds 4 \, 294 \, 967 \, 297\) | Sequence of Fermat Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds 641 \times 6 \, 700 \, 417\) | Prime Decomposition of $5$th Fermat Number |
Hence the result.
$\blacksquare$