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Difference between revisions of "Lifting the Exponent Lemma"
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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>):  
 
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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From (http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYy82LzdjNTI1OGIyMmNjYmZkZGY4MDhhY2ViZTc3MGE1NDRmMzFhMTEzLnBkZg==&rn=TGlmdGluZyBUaGUgRXhwb25lbnQgTGVtbWEgLSBBbWlyIEhvc3NlaW4gUGFydmFyZGkgLSBWZXJzaW9uIDMucGRm):
 
  
 
<math>\nu_p(x^n-y^n)=\nu_p(x-y)+\nu_p(n)</math>, if <math>p|x-y</math>.
 
<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.  
 
<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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From (https://arxiv.org/abs/1810.11456):
 
  
 
<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)=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(x+y)</math> if <math>2|x+y</math> and <math>n</math> is odd.
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== External Links ==
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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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Revision as of 13:04, 11 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 $p$ refer to an odd prime. We can split up LTE into six identities (where $\nu_p(Z)$ represents the largest factor of $p$ that divides $Z$):

$\nu_p(x^n-y^n)=\nu_p(x-y)+\nu_p(n)$, if $p|x-y$.

$\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(n),$ if $4|x-y$.

$\nu_2(x^n-y^n)=\nu_2(x-y)+\nu_2(x+y)+\nu_2(n)-1$, if $2|x-y$ and $n$ is even.

$\nu_p(x^n+y^n)=\nu_p(x+y)+\nu_p(n)$, if $p|x+y$ and $n$ is odd.

$\nu_2(x^n+y^n)=1$, 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.

External Links

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