Difference between revisions of "Sophie Germain Identity"
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=== Intermediate === | === Intermediate === | ||
*Compute <math>\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)}</math>. ([[1987 AIME Problems/Problem 14|1987 AIME, #14]]) | *Compute <math>\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)}</math>. ([[1987 AIME Problems/Problem 14|1987 AIME, #14]]) | ||
| − | *Find the largest prime divisor of <math>25^2+72^2</math>. ([[Mock AIME 5 2005-2006 Problems/ | + | *Find the largest prime divisor of <math>25^2+72^2</math>. ([[Mock AIME 5 2005-2006 Problems/Pro |
== See Also == | == See Also == | ||
Revision as of 07:55, 6 June 2010
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/Pro
. (
. ([[Mock AIME 5 2005-2006 Problems/Pro
