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

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TopNewestCould you please provide a proof for the sum you have claimed to be true? Thanks – Michael Stevenson · 1 year, 1 month ago

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