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Lifting the Exponent Lemma

Revision as of 19:52, 27 April 2025 by Btgames (talk | contribs) (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.

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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