Aurifeuillian Factorization/Examples/2^4n+2 + 1

Example of Aurifeuillian Factorization

$2^{4 n + 2} + 1 = \paren {2^{2 n + 1} - 2^{n + 1} + 1} \paren {2^{2 n + 1} + 2^{n + 1} + 1}$

Proof

From Sum of Squares as Product of Factors with Square Roots:

$x^2 + y^2 = \paren {x + \sqrt {2 x y} + y} \paren {x - \sqrt {2 x y} + y}$

Let $x = 2^{2 n + 1}$ and $y = 1$.

Then:

\(\ds 2^{4 n + 2} + 1\)

\(=\)

\(\ds \paren {2^{2 n + 1} }^2 + 1^2\)

\(\ds \)

\(=\)

\(\ds \paren {2^{2 n + 1} + \sqrt {2 \times \paren {2^{2 n + 1} } \times 1} + 1} \paren {2^{2 n + 1} - \sqrt {2 \times \paren {2^{2 n + 1} } \times 1} + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {2^{2 n + 1} + \sqrt {2^{2 n + 2} } + 1} \paren {2^{2 n + 1} - \sqrt {2^{2 n + 2} } + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {2^{2 n + 1} - 2^{n + 1} + 1} \paren {2^{2 n + 1} + 2^{n + 1} + 1}\)

$\blacksquare$

Historical Note

According to David Wells in his Curious and Interesting Numbers, 2nd ed. of $1997$, this identity was established by François Édouard Anatole Lucas, who generalised the result:

$2^{58} + 1 = \paren {2^{29} - 2^{15} + 1} \paren {2^{29} + 2^{15} + 1}$

which had been discovered by Léon-François-Antoine Aurifeuille in $1871$.

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $2^{58} + 1$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2^{58} + 1$