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Lifting the Exponent Lemma: Difference between revisions

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Lifting the exponent allows one to calculate the highest power of an integer that divides various numbers given certain information. It is extremely powerful and can sometimes "blow up" otherwise challenging problems.  
Lifting the exponent allows one to calculate the highest power of an integer that divides various numbers given certain information. It is extremely powerful and can sometimes "blow up" otherwise challenging problems.  


Let <math>p</math> refer to an odd prime. We can split up LTE into six identities (where <math>\nu_p(Z)</math> represents the largest factor of <math>p</math> that divides <math>Z</math>):  
Let <math>p</math> be a prime such that <math>p \nmid x</math> and <math>p \nmid y</math>. LTE comprises of the following identities (where <math>\nu_p(Z)</math> represents the largest factor of <math>p</math> that divides <math>Z</math>):


<math>\nu_p(x^n-y^n)=\nu_p(x-y)+\nu_p(n)</math>, if <math>p|x-y</math>.
* When <math>p</math> is odd:
 
** <math>\nu_p(x^n-y^n)=\nu_p(x-y)+\nu_p(n)</math>, if <math>p|x-y</math>.
<math>\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(n),</math> if <math>4|x-y</math>.  
** <math>\nu_p(x^n+y^n)=\nu_p(x+y)+\nu_p(n)</math>, if <math>p|x+y</math> and <math>n</math> is odd.
 
** <math>\nu_p(x^n+y^n)=0</math>, if <math>p|x+y</math> and <math>n</math> is even.
<math>\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(x+y)+\nu_2(n)-1</math>, if <math>2|x-y</math> and <math>n</math> is even.  
* When <math>p=2</math>:
 
** <math>\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(x+y)+\nu_2(n)-1</math>, if <math>2|x-y</math> and <math>n</math> is even.  
<math>\nu_p(x^n+y^n)=\nu_p(x+y)+\nu_p(n)</math>, if <math>p|x+y</math> and <math>n</math> is odd.  
** <math>\nu_2(x^n-y^n)=\nu(x-y)</math> if <math>2|x-y</math> and <math>n</math> is odd.
 
** Corollaries:
<math>\nu_2(x^n+y^n)=1</math>, if <math>2|x+y</math> and <math>n</math> is even.
*** <math>\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(n),</math> if <math>4|x-y</math>.
 
*** <math>\nu_2(x^n+y^n)=1</math>, if <math>2|x+y</math> and <math>n</math> is even.
<math>\nu_2(x^n+y^n)=\nu(x+y)</math> if <math>2|x+y</math> and <math>n</math> is odd.
*** <math>\nu_2(x^n+y^n)=\nu_2(x+y)</math>, if <math>2|x+y</math> and <math>n</math> is odd.


== External Links ==
== External Links ==

Revision as of 01:22, 26 March 2025

Lifting the exponent allows one to calculate the highest power of an integer that divides various numbers given certain information. It is extremely powerful and can sometimes "blow up" otherwise challenging problems.

Let $p$ be a prime such that $p \nmid x$ and $p \nmid y$. LTE comprises of the following identities (where $\nu_p(Z)$ represents the largest factor of $p$ that divides $Z$):

  • When $p$ is odd:
    • $\nu_p(x^n-y^n)=\nu_p(x-y)+\nu_p(n)$, if $p|x-y$.
    • $\nu_p(x^n+y^n)=\nu_p(x+y)+\nu_p(n)$, if $p|x+y$ and $n$ is odd.
    • $\nu_p(x^n+y^n)=0$, if $p|x+y$ and $n$ is even.
  • When $p=2$:
    • $\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(x+y)+\nu_2(n)-1$, if $2|x-y$ and $n$ is even.
    • $\nu_2(x^n-y^n)=\nu(x-y)$ if $2|x-y$ and $n$ is odd.
    • Corollaries:
      • $\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(n),$ if $4|x-y$.
      • $\nu_2(x^n+y^n)=1$, if $2|x+y$ and $n$ is even.
      • $\nu_2(x^n+y^n)=\nu_2(x+y)$, if $2|x+y$ and $n$ is odd.

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