Difference between revisions of "Sophie Germain Identity"

Revision as of 21:45, 25 May 2019

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, we begin by completing the square and then factor as a difference of squares:

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

Problems

Introductory

Prove that if $n>1$ then $n^4 + 4^n$ is composite. (1978 Kurschak Competition)

Intermediate

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)

@@ Line 12: / Line 12: @@
 === Introductory ===
-*Prove that if <math>n>1</math> then <math>n^4 + 4^n</math> is [[composite]].
+*Prove that if <math>n>1</math> then <math>n^4 + 4^n</math> is [[composite]]. (1978 Kurschak Competition)
 === Intermediate ===

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Difference between revisions of "Sophie Germain Identity"

Revision as of 21:45, 25 May 2019

Contents

Problems

Introductory

Intermediate

See Also