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/Pro | + | *Find the largest prime divisor of <math>25^2+72^2</math>. ([[Mock AIME 5 2005-2006 Problems/Pro]]) |
| + | *Calculate the value of <math>\dfrac{2014^4+4 \times 2013^4}{2013^2+4027^2}-\dfrac{2012^2+4 \times 2013^4}{2013^2+4025^2}</math>. ([[BMO 2013 #1]]) | ||
== See Also == | == See Also == | ||
Revision as of 00:04, 21 August 2014
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) - Calculate the value of
. (BMO 2013 #1)
. (
. (
. (
