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Difference between revisions of "Sophie Germain Identity"

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The '''Sophie Germain Identity''' states that:
 
The '''Sophie Germain Identity''' states that:
  
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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>
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<cmath>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</cmath>
  
 
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]]:
 
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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== Problems ==
 
== Problems ==
 
=== Introductory ===
 
=== Introductory ===
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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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*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|Source]])
  
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*Prove that if <math>n>0</math> then <math>n^2 + 4^n</math> is [[composite]]. (1978 Kurschak Competition)
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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 ===
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*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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*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|Source]])
  
 
*Find the largest prime divisor of <math>5^4+4 \cdot 6^4</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)
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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)
 
*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]])
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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|Source]])
  
 
== See Also ==
 
== See Also ==
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* [https://mathshistory.st-andrews.ac.uk/Biographies/Germain/ MacTutor biography of Sophie Germain]
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* [[Sophie Germain]]
  
 
[[Category:Algebra]]
 
[[Category:Algebra]]

Latest revision as of 16:05, 30 January 2025

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

  • What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$? (Source)
  • 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. (Source)

See Also

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