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Question Number 1557 by 123456 last updated on 19/Aug/15

find α, β, γ such that

α = β^{2}

β^{3} = γ^{4}

γ^{5} = α^{6}

Commented by 123456 last updated on 19/Aug/15

{\begin{cases} α = β^{2} \\ β^{3} = γ^{4} \\ γ^{5} = α^{6} \end{cases} \Rightarrow {\begin{cases} α^{15} = β^{30} \\ β^{30} = γ^{40} \\ γ^{40} = α^{48} \end{cases}

[(1, 2), (3, 4), (5, 6)]

[(3, 6), (6, 8), (5, 6)]

[(15, 30), (30, 40), (40, 48)]

α^{15} = α^{48} \Rightarrow α^{15} (α^{33} - 1) = 0

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

α = β^{2} \Rightarrow α^{3} = β^{6} \dots \dots \dots \dots . . (I)

β^{3} = γ^{4} \Rightarrow β^{6} = γ^{8} \dots \dots \dots \dots (I I)

F r o m (I) a n d (I I) α^{3} = γ^{8} \Rightarrow α^{6} = γ^{16} \dots . . (I I I)

γ^{5} = α^{6} \Rightarrow α^{6} = γ^{5} \dots \dots \dots \dots \dots \dots \dots \dots . (I V)

F r o m (I I I) a n d (I V)

γ^{5} = γ^{16} \Rightarrow γ^{16} - γ^{5} = 0 \Rightarrow γ^{5} (γ^{11} - 1) = 0

γ = 0 \lor γ^{11} = 1, γ i s 11 t h r o o t o f u n i t y .

[A b o v e w a s t h e p r o c e s s o f e l i m i n a t i n g α a n d β .

I n t h e s a m e w a y b y e l i m i n a t i n g β a n d γ a n d a f t e r t h a t

b y α a n d γ w e w i l l g e t r e s u l t s a b o u t α a n d β

r e s p e c t i v e l y .]

R e s u l t s a b o u t α a n d β

α = 0 \lor α^{33} = 1, α i s 33 r d r o o t o f u n i t y .

β = 0 \lor β^{33} = 1, β i s 33 r d r o o t o f u n i t y .

(α, β, γ) = (0, 0, 0)

S o l u t i o n o f α o r β : {0} \cup {1, w, ω^{2}, \dots ω^{32}}, ω = c o s \frac{2 π}{33} + ı s i n \frac{2 π}{33}

S o l u t i o n o f γ : {0} \cup {1, ρ, ρ^{2}, \dots ρ^{10}}, ρ = c o s \frac{2 π}{11} + ı s i n \frac{2 π}{11}

L e t (α, β, γ) = (ω^{p}, ω^{q}, ρ^{r}) f u l f i l l i n g g i v e n c o n d i t i o n s .

w^{p} = {(ω^{q})}^{2} \Rightarrow p = 2 q \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots . . I

{(ω^{q})}^{3} = {(ρ^{r})}^{4} \Rightarrow ω^{3 q} = ρ^{4 r} \Rightarrow ω^{15 q} = ρ^{20 r} \dots \dots \dots . . I I

{(ρ^{r})}^{5} = {(ω^{p})}^{6} \Rightarrow ρ^{5 r} = ω^{6 p} \Rightarrow ρ^{20 r} = ω^{24 p} \dots \dots \dots . I I I

p = 2 q \dots \dots . [f r o m I]

24 p = 15 q \Rightarrow 8 p = 5 q \dots \dots \dots \dots \dots [f r o m I I a n d I I I]

8 (2 q) = 5 q \Rightarrow q = 0 \Rightarrow p = 2 (0) = 0 \Rightarrow p = 0

ρ^{20 r} = ω^{24 p} = ω^{24 (0)} = 1 = ρ^{0} \Rightarrow 20 r = 0 \Rightarrow r = 0

S o l . S e t : {(0, 0, 0), (1, 1, 1)}