Difference of Powers

For integers a,b,m,na, b, m, n, prove that gcd(ambm,anbn)=agcd(m,n)bgcd(m,n).gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = {a}^{gcd(m,n)} - {b}^{gcd(m,n)}.

Solution

Let d=gcd(m,n)d = gcd(m,n), thus dmd|m and dnd|n. We then let m=pdm=pd and n=pdn=pd.

Hence gcd(ambm,anbn)=gcd(apdbpd,aqdbqd).gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = gcd({a}^{pd} - {b}^{pd},{a}^{qd} - {b}^{qd}).

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

gcd(ambm,anbn)=gcd(ApBp,AqBq).gcd({a}^{m} - {b}^{m}, {a}^{n} - {b}^{n}) = gcd({A}^{p} - {B}^{p},{A}^{q} - {B}^{q}).

Since ApBpAB=k=0p1Apk1Bk\frac{{A}^{p} - {B}^{p}}{A-B} = \sum _{ k=0 }^{ p-1 }{ { A }^{ p-k-1 }{ B }^{ k } } and AqBqAB=k=0q1Aqk1Bk\frac{{A}^{q} - {B}^{q}}{A-B} = \sum _{ k=0 }^{ q-1 }{ { A }^{ q-k-1 }{ B }^{ k } } we get

gcd(ApBp,AqBq)=AB.gcd({A}^{p} - {B}^{p},{A}^{q} - {B}^{q}) = A-B.

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

gcd(ambm,anbn)=agcd(m,n)bgcd(m,n).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
4 years, 11 months ago

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

Michael Stevenson - 3 years, 3 months ago

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