Question Number 1498 by Rasheed Soomro last updated on 14/Aug/15

Find complex numbers α and β such that α^( m) =β^( n) and β^( m) =α^n , m,n ∈ Z Determine formula for the number of pairs (α,β) fulfilling the above conditions. ( You may ignore this part in your answer)

Commented by123456 last updated on 14/Aug/15

(n,m)≠(0,0)⇒(α,β)=(0,0) is one of the solutions

Commented byRasheed Ahmad last updated on 14/Aug/15

An other pair is (1,1) but these are real solutions only.

Commented by123456 last updated on 15/Aug/15

{ ((α^m =β^n )),((β^m =α^n )) :}⇒_(β→α) ^(α→β) { ((β^m =α^n )),((α^m =β^n )) :} (α,β) is a solution, then (β,α) is also a solution

Commented byRasheed Ahmad last updated on 15/Aug/15

Good deduction!

Commented by123456 last updated on 15/Aug/15

(n,m)∈(N^∗ )^2 ,p≤max(rn,rm)+1 k=lcm(n,m) { ((α^n =β^m )),((β^n =α^m )) :}⇒ { ((α^(nk_1 ) =β^(mk_1 ) )),((β^(nk_2 ) =α^(mk_2 ) )) :} mk_1 =nk_2 =k α^(nk_1 ) −α^(mk_2 ) =0⇒α^r_1 (α^(r_2 −r_1 ) −1)=0 r_1 =min(nk_1 ,mk_2 ),r_2 =max(nk_1 ,mk_2 ) α^r_1 =0∨α^r =1 r=r_2 −r_1

Commented byRasheed Soomro last updated on 15/Aug/15

I have used your this result in my answer to Q 1497

Answered by Rasheed Soomro last updated on 17/Aug/15

Let gcd(m,n)=k ⇒ lcm(m,n)=((mn)/k) α^m =β^( n) ⇒( α^m )^(m/k) =(β^( n) )^(m/k) ⇒ α^(m^2 /k) =β^((mn)/k) (I) β^( m) =α^( n) ⇒ (β^( m) )^(n/k) = (α^n )^(n/k) ⇒ β^((mn)/k) =α^(n^2 /k) (II) Our goal is to achieve β ′s common exponent ( of course least common exponent (((mn)/k) ) is suitable for simplified solution) in order to eliminate β in above steps. From (I) and (II) α^(m^2 /k) = α^(n^2 /k) As m and n are exchangeable here we may assume m≥n (If n>m then it will be merely matter of exchanging symbols) α^(m^2 /k) − α^(n^2 /k) =0 ⇒ α^(n^2 /k) ( α^((m^2 /k) − (n^2 /k)) − 1)=0 ⇒ α^(n^2 /k) = 0 ∨ α^((m^2 − n^2 )/k) =1 ⇒ α=0 ∨ α=^((m^2 − n^2 )/k) (√( 1)) (α is (((m^2 − n^2 )/k))th root of unity) α=0 ⇒β=0 [ No need to show process ] ∴ α is (((m^2 − n^2 )/k))th root of unity. Similarly it can be shown that, β is also (((m^2 − n^2 )/k))th root of unity. nth roots of unity are given by: ( cos ((2π)/n) +ı sin ((2π)/n) )^j j=0,1,2,....,n−1 Hence, (((m^2 − n^2 )/k)) th roots of unity will be: ( cos ((2πk)/(m^2 −n^2 )) +ı sin ((2πk)/(m^2 −n^2 )) )^j j=0,1,2,...(.((m^2 − n^2 )/k) −1) Let cos ((2πk)/(m^2 −n^2 )) +ı sin ((2πk)/(m^2 −n^2 )) =ω (((m^2 − n^2 )/k)) th roots of unity=1,ω,ω^2 ,......,ω^((m^2 − n^2 −k)/k) [ Let ((m^2 −n^2 )/k) = N ] mth power of ∽ =1, ω^( m(mod N)) ,ω^( 2m(mod N)) ,...ω^( m(N−1)(mod N)) nth power of ∽ =1, ω^( n(mod N)) ,ω^( 2n(mod N)) ,...ω^( n(N−1)(mod N)) Let j_1 , j_2 =0,1,2,...(.((m^2 − n^2 )/k) −1) Let α = ω^( j_1 ) and β = ω^( j_2 ) are two Nth roots such that α^m =β^( n) ⇒ ω^( mj_1 (mod N)) =ω^( nj_2 (mod N)) ⇒mj_1 (mod N)=nj_2 (mod N) α^n =β^( m) ⇒ ω^( nj_1 (mod N)) =ω^( mj_2 (mod N)) ⇒nj_1 (mod N)=mj_2 (mod N) Therefore −−−−−−−−−−−−−−−−−−−−−−−−− For mj_1 (mod ((m^2 −n^2 )/k) )=nj_2 (mod ((m^2 −n^2 )/k) ) and nj_1 (mod ((m^2 −n^2 )/k) )=mj_2 (mod ((m^2 −n^2 )/k) ) ( α, β)=(ω^j_1 , ω^j_2 )=(ω^j_2 , ω^j_1 ) where ω is (((m^2 − n^2 )/k)) th root of unity( with m≥n).

Answered by Rasheed Soomro last updated on 17/Aug/15

Formula for number of pairs (𝛂,𝛃) fulfilling the conditions given in the question. ((m^2 −n^2 +k)/k) Since the number of (((m^2 −n^2 )/k))th roots is ((m^2 −n^2 )/k) and any one of them may be the value of α and to every α one value of β is attached. Hence number of (α,β) pairs is ((m^2 −n^2 )/k) . But (0,0) also satisfy the conditions (and this is not included in the roots) So ((m^2 −n^2 )/k)+1=((m^2 −n^2 +k)/k) .Remember that k=gcd(m,n) and m≥n.

Commented byRasheed Soomro last updated on 20/Aug/15

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