Difference between revisions of "Sophie Germain Identity"

Revision as of 12:11, 28 December 2016

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:

$\begin{align*} a^4 + 4b^4 &= a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\ &= (a^2 + 2b^2)^2 - (2ab)^2 \\ &= (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab) \end{align*}$

Compute $\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}$ . (1987 AIME, #14)

Find the largest prime divisor of $5^4+4 \cdot 6^4$ . (Mock AIME 5 2005-2006 Problems/Pro)

Calculate the value of $\dfrac{2014^4+4 \times 2013^4}{2013^2+4027^2}-\dfrac{2012^4+4 \times 2013^4}{2013^2+4025^2}$ . (BMO 2013 #1)

@@ Line 17: / Line 17: @@
 *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>5^4+4 \cdot 6^4</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{2012^4+4 \times 2013^4}{2013^2+4025^2}</math>. (BMO 2013 #1)