# Difference of Powers

For integers $$a, b, m, n$$, prove that $gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = {a}^{gcd(m,n)} - {b}^{gcd(m,n)}.$

Solution

Let $d = gcd(m,n)$, thus $d|m$ and $d|n$. We then let $m=pd$ and $n=pd$.

Hence $gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = gcd({a}^{pd} - {b}^{pd},{a}^{qd} - {b}^{qd}).$

Let $A ={a}^{d}$ and $B = {b}^{d}$, it follows that

$gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = gcd({A}^{p} - {B}^{p},{A}^{q} - {B}^{q}).$

Since $\frac{{A}^{p} - {B}^{p}}{A-B} = \sum _{ k=0 }^{ p-1 }{ { A }^{ p-k-1 }{ B }^{ k } }$ and $\frac{{A}^{q} - {B}^{q}}{A-B} = \sum _{ k=0 }^{ q-1 }{ { A }^{ q-k-1 }{ B }^{ k } }$ we get

$gcd({A}^{p} - {B}^{p},{A}^{q} - {B}^{q}) = A-B.$

Since $A-B = {a}^{d} - {b}^{d}$ and $d =gcd(m,n)$, we prove that

$gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = {a}^{gcd(m,n)} - {b}^{gcd(m,n)}.$

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
7 years ago

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Could you please provide a proof for the sum you have claimed to be true? Thanks

- 5 years, 4 months ago