Difference between revisions of "Sophie Germain Identity"
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<div style="text-align:center;"><math>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</math></div> | <div style="text-align:center;"><math>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</math></div> | ||
| − | One can prove this identity simply by multiplying out the right side and verifying that it equals the left. To derive the [[ | + | 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]]: |
<math>\begin{align*}a^4 + 4b^4 & = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\ | <math>\begin{align*}a^4 + 4b^4 & = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\ | ||
Revision as of 20:52, 12 March 2009
The Sophie Germain Identity states that:
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:
$\begin{align*}a^4 + 4b^4 & = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\ & = (a^2 + 2b^2)^2 - (2ab)^2 \\ & = (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Problems
Introductory
a 
then 
is Intermediate
- Compute
. (1987 AIME, #14) - Find the largest prime divisor of
. (Mock AIME 5 2005-2006 Problems/Problem 5)
. (
. (
