Find-complex-numbers-and-such-that-m-n-and-m-n-m-n-Z-Determine-formula-for-the-number-of-pairs-fulfilling-the-above-conditions-You-may-ignore-this-part

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

Find complex numbers α a n d β such that

α^{m} = β^{n} a n d β^{m} = α^{n}, m, n \in Z

D etermine formula for the number of pairs (α, β) fulfilling the

above conditions . (You may ignore this part in your answer)

Commented by 123456 last updated on 14/Aug/15

(n, m) \neq (0, 0) \Rightarrow (α, β) = (0, 0) is one of the solutions

Commented by Rasheed Ahmad last updated on 14/Aug/15

An other pair is (1, 1) b u t t h e s e a r e

r e a l s o l u t i o n s o n l y .

Commented by 123456 last updated on 15/Aug/15

{\begin{cases} α^{m} = β^{n} \\ β^{m} = α^{n} \end{cases} \underset{β \to α}{\overset{α \to β}{\Rightarrow}} {\begin{cases} β^{m} = α^{n} \\ α^{m} = β^{n} \end{cases}

(α, β) is a solution, then (β, α) is also a solution

Commented by Rasheed Ahmad last updated on 15/Aug/15

Good deduction!

Commented by 123456 last updated on 15/Aug/15

(n, m) \in {(N^{*})}^{2}, p ⩽ \max (r n, r m) + 1

k = lcm (n, m)

{\begin{cases} α^{n} = β^{m} \\ β^{n} = α^{m} \end{cases} \Rightarrow {\begin{cases} α^{{n k}_{1}} = β^{{m k}_{1}} \\ β^{{n k}_{2}} = α^{{m k}_{2}} \end{cases}

{m k}_{1} = {n k}_{2} = k

α^{{n k}_{1}} - α^{{m k}_{2}} = 0 \Rightarrow α^{r_{1}} (α^{r_{2} - r_{1}} - 1) = 0

r_{1} = \min ({n k}_{1}, {m k}_{2}), r_{2} = \max ({n k}_{1}, {m k}_{2})

α^{r_{1}} = 0 \lor α^{r} = 1

r = r_{2} - r_{1}

Commented by Rasheed Soomro last updated on 15/Aug/15

I h a v e u s e d y o u r t h i s r e s u l t i n m y a n s w e r t o Q 1497

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

Let \gcd (m, n) = k \Rightarrow lcm (m, n) = \frac{mn}{k}

α^{m} = β^{n} \Rightarrow {(α^{m})}^{\frac{m}{k}} = {(β^{n})}^{\frac{m}{k}} \Rightarrow α^{\frac{m^{2}}{k}} = β^{\frac{mn}{k}} (I)

β^{m} = α^{n} \Rightarrow {(β^{m})}^{\frac{n}{k}} = {(α^{n})}^{\frac{n}{k}} \Rightarrow β^{\frac{mn}{k}} = α^{\frac{n^{2}}{k}} (II)

Our goal is to achieve β^{'} s common exponent

(of course least common exponent (\frac{mn}{k}) is suitable for simplified solution)

in order to eliminate β in above steps .

From (I) and (II)

α^{\frac{m^{2}}{k}} = α^{\frac{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^{\frac{m^{2}}{k}} - α^{\frac{n^{2}}{k}} = 0 \Rightarrow α^{\frac{n^{2}}{k}} (α^{\frac{m^{2}}{k} - \frac{n^{2}}{k}} - 1) = 0

\Rightarrow α^{\frac{n^{2}}{k}} = 0 \lor α^{\frac{m^{2} - n^{2}}{k}} = 1

\Rightarrow α = 0 \lor α =^{\frac{m^{2} - n^{2}}{k}} \sqrt{1} (α is (\frac{m^{2} - n^{2}}{k}) th root of unity)

α = 0 \Rightarrow β = 0 [No need to show process]

∴ α is (\frac{m^{2} - n^{2}}{k}) th root of unity .

Similarly it can be shown that,

β is also (\frac{m^{2} - n^{2}}{k}) th root of unity .

nth roots of unity are given by :

{(\cos \frac{2 π}{n} + ı \sin \frac{2 π}{n})}^{j} j = 0, 1, 2, \dots ., n - 1

Hence,

(\frac{m^{2} - n^{2}}{k}) t h roots of unity will be :

{(\cos \frac{2 π k}{m^{2} - n^{2}} + ı \sin \frac{2 π k}{m^{2} - n^{2}})}^{j} j = 0, 1, 2, \dots (. \frac{m^{2} - n^{2}}{k} - 1)

Let \cos \frac{2 π k}{m^{2} - n^{2}} + ı \sin \frac{2 π k}{m^{2} - n^{2}} = ω

(\frac{m^{2} - n^{2}}{k}) t h roots of unity = 1, ω, ω^{2}, \dots \dots, ω^{\frac{m^{2} - n^{2} - k}{k}}

[Let \frac{m^{2} - n^{2}}{k} = N]

m t h power of ∽ = 1, ω^{m (\mod N)}, ω^{2 m (\mod N)}, \dots ω^{m (N - 1) (\mod N)}

n t h power of ∽ = 1, ω^{n (\mod N)}, ω^{2 n (\mod N)}, \dots ω^{n (N - 1) (\mod N)}

Let j_{1}, j_{2} = 0, 1, 2, \dots (. \frac{m^{2} - n^{2}}{k} - 1)

Let α = ω^{j_{1}} and β = ω^{j_{2}} are two N t h roots such that

α^{m} = β^{n} \Rightarrow ω^{{mj}_{1} (\mod N)} = ω^{{nj}_{2} (\mod N)} \Rightarrow {mj}_{1} (\mod N) = {nj}_{2} (\mod N)

α^{n} = β^{m} \Rightarrow ω^{{nj}_{1} (\mod N)} = ω^{{mj}_{2} (\mod N)} \Rightarrow {nj}_{1} (\mod N) = {mj}_{2} (\mod N)

Therefore

- - - - - - - - - - - - - - - - - - - - - - - - -

For {mj}_{1} (\mod \frac{m^{2} - n^{2}}{k}) = {nj}_{2} (\mod \frac{m^{2} - n^{2}}{k})

and {nj}_{1} (\mod \frac{m^{2} - n^{2}}{k}) = {mj}_{2} (\mod \frac{m^{2} - n^{2}}{k})

(α, β) = (ω^{j_{1}}, ω^{j_{2}}) = (ω^{j_{2}}, ω^{j_{1}})

where ω is (\frac{m^{2} - n^{2}}{k}) t h 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 .

\frac{m^{2} - n^{2} + k}{k}

S ince the number of (\frac{m^{2} - n^{2}}{k}) t h roots is \frac{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 \frac{m^{2} - n^{2}}{k} . But (0, 0) also satisfy the

conditions (and this is not included in the roots)

So \frac{m^{2} - n^{2}}{k} + 1 = \frac{m^{2} - n^{2} + k}{k} . Remember that k = \gcd (m, n) and

m ⩾ n .

Commented by Rasheed Soomro last updated on 20/Aug/15

W h o d i d l i k e ? S u p p o s e l i k i n g b y a b c i s m o r e i m p o r t a n t

f o r m e t h e n I w o u l d l i k e t o k n o w a b o u t {l i k e r}^{'} s I D .

S o l i k i n g s h o u l d b e r e c o r d e d w i t h I D .

PL demand for this .

O f c o u r s e {L i k e r s}^{'} {I D}^{'} s {s h o u l d n}^{'} t b e t h e r e f o r f u l l

t i m e b u t w h e n w e c l i c k \dots \dots . W e m u s t k n o w,^{″} W h o

d i d l i k e ?^{″}