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Difference between revisions of "Sophie Germain Identity"
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<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>
  
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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 [[completing the square]] and then factor as a [[difference of squares]]:
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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]]:
  
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<math>\begin{align*}a^4 + 4b^4 & = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\
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<cmath>\begin{align*}
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& = (a^2 + 2b^2)^2 - (2ab)^2 \\
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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)
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& = (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)\end{align*}</math>
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\end{align*}</cmath>
  
 
== Problems ==
 
== Problems ==
 
=== Introductory ===
 
=== Introductory ===
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*Is <math>4^{545} + 545^{4}</math> a [[prime]]?
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*What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>? ([[2020 AMC 10B Problems/Problem 22|2020 AMC 10B, #22]])
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*Prove that if <math>n>1</math> then <math>n^4 + 4^n</math> is [[composite]].
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*Prove that if <math>n>1</math> then <math>n^4 + 4^n</math> is [[composite]]. (1978 Kurschak Competition)
  
 
=== 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]])
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*Find the largest prime divisor of <math>25^2+72^2</math>. ([[Mock AIME 5 2005-2006 Problems/Problem 5]])
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*Find the largest prime divisor of <math>5^4+4 \cdot 6^4</math>. (Mock AIME 5 2005-2006 Problems/Pro)
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*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)
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*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)
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*Prove that there exist infinite natural numbers <math>m</math> fulfilling the following property: For all natural numbers <math>n</math>, <math>n^4+m</math> is not a prime number. ([[1969 IMO Problems/Problem 1|IMO 1969 #1]])
  
 
== See Also ==
 
== See Also ==
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* [http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Germain.html MacTutor biography of Sophie Germain]
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* [https://mathshistory.st-andrews.ac.uk/Biographies/Germain/ MacTutor biography of Sophie Germain]
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[[Category:Algebra]]

Latest revision as of 18:11, 4 August 2024

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

  • 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)
  • Prove that there exist infinite natural numbers $m$ fulfilling the following property: For all natural numbers $n$, $n^4+m$ is not a prime number. (IMO 1969 #1)

See Also

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