📂Number Theory

Proof of Exponential Auxiliary Lemma

Theorem

$n \in \mathbb{N}$, $x , y \in \mathbb{Z}$, prime numbers $p \ne 2$ satisfy $$ \gcd (n,p) = 1 \\ p \mid (x - y) \\ p \nmid x \\ p \nmid y $$ then $$ v_{p} \left( x^{n} - y^{n} \right) = v_{p} \left( x - y \right) + v_{p} (n) $$

• $v_{p} (a)$ represents the $p$-ary expansion of $a$.

Proof ¹

Strategy: It naturally follows from the properties of $p$-ary expansion. However, proving those properties is quite lengthy. The key is to present those properties first, which can be proved with elementary number theory knowledge, so don’t be intimidated and give it a try.

Lemma (Properties of $p$-ary):

• [3]: $n \in \mathbb{N}$, $x , y \in \mathbb{Z}$, prime numbers $p$ satisfy $$ \gcd (n,p) = 1 \\ p \mid (x \mp y) \\ p \nmid x \\ p \nmid y $$ then $$ v_{p} \left( x^{n} \pm y^{n} \right) = v_{p} \left( x \pm y \right) $$
• [4]: $x , y \in \mathbb{Z}$, prime numbers $p \ne 2$ satisfy $$ p \mid (x - y) \\ p \nmid x \\ p \nmid y $$ then $$ v_{p} \left( x^{p} - y^{p} \right) = v_{p} \left( x - y \right) +1 $$

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$$ n := p^{\alpha} b \\ \gcd (p,b) = 1 $$ By choosing a natural number $n, b , \alpha$ that satisfies the above two conditions, we get $v_{p} (n) = \alpha$. Then, by using the lemma [3], $$ \begin{align*} v_{p} \left( x^{n} - y^{n} \right) =& v_{p} \left( \left[ x^{p^{\alpha}} \right]^{b} - \left[ y^{p^{\alpha}} \right]^{b} \right) \\ =& v_{p} \left( x^{p^{\alpha}} - y^{p^{\alpha}} \right) \\ =& v_{p} \left( \left[ x^{p^{\alpha - 1}} \right]^{p} - \left[ y^{p^{\alpha - 1}} \right]^{p} \right) \end{align*} $$ Solving recursively using lemma [4] gives $$ \begin{align*} v_{p} \left( \left[ x^{p^{\alpha - 1}} \right]^{p} - \left[ y^{p^{\alpha - 1}} \right]^{p} \right) =& v_{p} \left( \left[ x^{p^{\alpha - 1}} \right] - \left[ y^{p^{\alpha - 1}} \right] \right) + 1 \\ =& v_{p} \left( \left[ x^{p^{\alpha - 2}} \right]^{p} - \left[ y^{p^{\alpha - 2}} \right]^{p} \right) + 1 \\ =& v_{p} \left( \left[ x^{p^{\alpha - 2}} \right] - \left[ y^{p^{\alpha - 2}} \right] \right) + 2 \\ &\vdots& \\ =& v_{p}(x-y) + \alpha \\ =& v_{p}(x-y) + v_{p} (n) \end{align*} $$ In conclusion, we obtain the following. $$ v_{p} \left( x^{n} - y^{n} \right) = v_{p} \left( x - y \right) + v_{p} (n) $$

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