Difference between revisions of "Sophie Germain Identity"

Revision as of 17:05, 4 February 2008

The Sophie Germain Identity, credited to Marie-Sophie Germain, states that:

$a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$

$a^4 + 4b^4 = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2$
$= (a^2 + 2b^2)^2 - 4a^2b^2$
$= (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)$

Compute $\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}$ . (1987 AIME, #14)

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 The proof involves [[completing the square]] and then [[difference of squares]].
-<div style="text-align:center;"><math>\displaystyle a^4 + 4b^4 = a^4 + 4a^2b^2 + 4b^4 - 4a^4b^4</math><br /><math>= (a^2 + 2b^2)^2 - 4a^4b^4</math><br /><math>\displaystyle = (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)</math></div>
+<div style="text-align:center;"><math>a^4 + 4b^4 = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2</math><br /><math>= (a^2 + 2b^2)^2 - 4a^2b^2</math><br /><math>= (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)</math></div>
 == Problems ==