Difference between revisions of "Sophie Germain Identity"
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| − | The '''Sophie Germain Identity''' | + | The '''Sophie Germain Identity''' states that: |
<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 [[factorization | factoring]], first [[complete the square | 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 \\ | |
| + | & = (a^2 + 2b^2)^2 - (2ab)^2 \\ | ||
| + | & = (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)\end{align*}</math> | ||
== Problems == | == Problems == | ||
=== Introductory === | === Introductory === | ||
*Is <math>4^{545} + 545^{4}</math> a [[prime]]? | *Is <math>4^{545} + 545^{4}</math> a [[prime]]? | ||
| − | *Prove that if <math>n>1</math> | + | *Prove that if <math>n>1</math> then <math>n^4 + 4^n</math> is [[composite]]. |
=== 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]]) | ||
| + | |||
| + | |||
| + | == See Also == | ||
| + | * [[http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Germain.html MacTutor biography of Sophie Germain]] | ||
Revision as of 18:52, 5 February 2008
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)
. (
