Difference between revisions of "Sophie Germain Identity"

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One can prove this identity simply by multiplying out the right side and verifying that it equals the left.  To derive the [[factoring]], first [[completing the square]] and then factor as a [[difference of squares]]:
 
One can prove this identity simply by multiplying out the right side and verifying that it equals the left.  To derive the [[factoring]], first [[completing the square]] and then factor as a [[difference of squares]]:
  
\begin{align*}a^4 + 4b^4 & = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\
+
<cmath>a^4 + 4b^4 = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\ \\ = (a^2 + 2b^2)^2 - (2ab)^2 \\ \\ = (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)</cmath>
& = (a^2 + 2b^2)^2 - (2ab)^2 \\
 
& = (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)\end{align*}
 
  
 
== Problems ==
 
== Problems ==

Revision as of 09:02, 3 May 2015

The Sophie Germain Identity states that:

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

One can prove this identity simply by multiplying out the right side and verifying that it equals the left. To derive the factoring, first completing the square and then factor as a difference of squares:

\[a^4 + 4b^4 = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\ \\ = (a^2 + 2b^2)^2 - (2ab)^2 \\ \\ = (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)\]

Problems

Introductory

Intermediate

See Also

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