Difference between revisions of "Sophie Germain Identity"

Revision as of 20:38, 31 December 2017

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)

Find the largest prime factor of $13^4+16^5-172^2$ , given that it is the product of three distinct primes. (ARML 2016 Individual #10)

For integers $n > 1$ , show that $n^4+4^n$ is never prime. (1978 Kurschak Competition)

@@ Line 22: / Line 22: @@
 *Find the largest prime factor of <math>13^4+16^5-172^2</math>, given that it is the product of three distinct primes. (ARML 2016 Individual #10)
+*For integers <math>n > 1</math>, show that <math>n^4+4^n</math> is never prime. (1978 Kurschak Competition)
 == See Also ==