Sophie Germain's Identity

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Theorem

For any two numbers $x$ and $y$:

$x^4 + 4 y^4 = \paren {x^2 + 2 y^2 + 2 x y} \paren {x^2 + 2 y^2 - 2 x y}$


Proof 1

\(\ds \) \(\) \(\ds \paren {x^2 + 2 y^2 + 2 x y} \paren {x^2 + 2 y^2 - 2 x y}\)
\(\ds \) \(=\) \(\ds x^4 + x^2 \cdot 2 y^2 - x^2 \cdot 2 x y + x^2 \cdot 2 y^2 + 4 y^2 - 2 y^2 \cdot 2 x y + x^2 \cdot 2 x y + 2 y^2 \cdot 2 x y - 2 x y \cdot 2 x y\)
\(\ds \) \(=\) \(\ds x^4 + 4 y^4\) gathering up terms and cancelling

$\blacksquare$


Proof 2

\(\ds \) \(\) \(\ds \paren {x^2 + 2 y^2 + 2 x y} \paren {x^2 + 2 y^2 - 2 x y}\)
\(\ds \) \(=\) \(\ds \paren {x^2 + 2 y^2}^2 - \paren {2 x y}^2\) Difference of Two Squares
\(\ds \) \(=\) \(\ds x^4 + 2 \cdot x^2 \cdot 2 y^2 + 4 y^2 - 2 x y \cdot 2 x y\)
\(\ds \) \(=\) \(\ds x^4 + 4 y^4\) gathering up terms and cancelling

$\blacksquare$


Source of Name

This entry was named for Marie-Sophie Germain.


Sources