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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 2$: Special Products and Factors: $2.19$
- 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 2$: Special Products and Factors: $2.19.$