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Difference between revisions of "Sophie Germain Identity"

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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 AMC10 B)
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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]])
  
 
*Prove that if <math>n>1</math> then <math>n^4 + 4^n</math> is [[composite]]. (1978 Kurschak Competition)
 
*Prove that if <math>n>1</math> then <math>n^4 + 4^n</math> is [[composite]]. (1978 Kurschak Competition)

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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