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Difference between revisions of "Lifting the Exponent Lemma"

(fixed incorrect first LTE identity.)
m (Changes V_p represents the largest *factor* to largest *power*)
 
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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.  
  
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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>):  
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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 power of <math>p</math> that divides <math>Z</math>):
  
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From (http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYy82LzdjNTI1OGIyMmNjYmZkZGY4MDhhY2ViZTc3MGE1NDRmMzFhMTEzLnBkZg==&rn=TGlmdGluZyBUaGUgRXhwb25lbnQgTGVtbWEgLSBBbWlyIEhvc3NlaW4gUGFydmFyZGkgLSBWZXJzaW9uIDMucGRm):
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* When <math>p</math> is odd:
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** <math>\nu_p(x^n-y^n)=\nu_p(x-y)+\nu_p(n)</math>, if <math>p|x-y</math>.
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** <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.
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** <math>\nu_p(x^n+y^n)=0</math>, if <math>p|x+y</math> and <math>n</math> is even.
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* When <math>p=2</math>:
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** <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.  
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** <math>\nu_2(x^n-y^n)=\nu(x-y)</math> if <math>2|x-y</math> and <math>n</math> is odd.
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** Corollaries:
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*** <math>\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(n),</math> if <math>4|x-y</math>.
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*** <math>\nu_2(x^n+y^n)=1</math>, if <math>2|x+y</math> and <math>n</math> is even.
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*** <math>\nu_2(x^n+y^n)=\nu_2(x+y)</math>, if <math>2|x+y</math> and <math>n</math> is odd.
  
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<math>\nu_p(x^n-y^n)=\nu_p(x-y)+\nu_p(n)</math>, if <math>p|x-y</math>.
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== External Links ==
  
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<math>\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(n),</math> if <math>4|x-y</math>.  
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* [//arxiv.org/abs/1810.11456]
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* [//services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYy82LzdjNTI1OGIyMmNjYmZkZGY4MDhhY2ViZTc3MGE1NDRmMzFhMTEzLnBkZg==&rn=TGlmdGluZyBUaGUgRXhwb25lbnQgTGVtbWEgLSBBbWlyIEhvc3NlaW4gUGFydmFyZGkgLSBWZXJzaW9uIDMucGRm]
  
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<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>.
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{{stub}}
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<math>\nu_p(x^n+y^n)=\nu_p(x+y)+\nu_p(n)</math>, if <math>p|x+y</math>.
 
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From (https://arxiv.org/abs/1810.11456):
 
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<math>\nu_2(x^n+y^n)=1</math>, if <math>2|x+y</math> and <math>n</math> is even.
 
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<math>\nu_2(x^n+y^n)=\nu(x+y)</math> if <math>2|x+y</math> and <math>n</math> is odd.
 

Latest revision as of 19:52, 27 April 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 power 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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