Analyze-for-integer-solution-a-b-c-are-fixed-positive-integers-ax-a-by-b-cz-c-

Question Number 4266 by Rasheed Soomro last updated on 06/Jan/16

Analyze for integer solution :

a, b, c are fixed positive integers .

{ax}^{a} + {by}^{b} = {cz}^{c}

Commented by Yozzii last updated on 06/Jan/16

I f (x, y, z) = (n^{1 / a}, j^{1 / b}, k^{1 / c}), w h e r e

n, j a n d k a r e t h e a - t h, b - t h & c - t h

p o w e r s o f p o s i t i v e i n t e g e r s, r e s p e c t i v e l y, t h e n

t h e e q u a t i o n {a x}^{a} + {b y}^{b} = {c z}^{c} r e d u c e s t o

a n + b j = c k .

a n + b j - c k = 0

\Rightarrow \frac{a}{c} n + \frac{b}{c} j = k

Commented by prakash jain last updated on 06/Jan/16

a = b = c = 1

x + y = z

a = b = c = 2

p y t h o g o r e a n t r i p l e s

a = b = c ⩾ 3 no solution

Commented by Rasheed Soomro last updated on 13/Jan/16

{ax}^{a} + {by}^{b} = {cz}^{c}

F o r a = 1, b = 2 and c = 3

x + 2 y^{2} = 3 z^{3}

x = 3 z^{3} - 2 y^{2}

T a k e v a l u e s o f y a n d z s o t h a t

3 z^{3} > 2 y^{2}

F o r e x a m p l e

x = 6, y = 3, z = 2

(6 + 2 {(3)}^{2} = 3 {(2)}^{3}

6 + 18 = 24

24 = 24)

F o r a = 2, b = 3 a n d c = 4

2 x^{2} + 3 y^{3} = 4 z^{4}

2 x^{2} + 3 y^{3} \in E \land 2 x^{2} \in E

\Rightarrow 3 y^{3} \in E \Rightarrow y \in E

F o r y = 0

2 x^{2} = 4 z^{4}

\frac{x^{2}}{z^{4}} = \frac{4}{2} = 2

F o r y = 0 n o i n t e g e r x, z

F o r y = 2

2 x^{2} + 24 = 4 z^{4}

2 z^{4} - x^{2} = 12

T h e a b o v e e q u a t i o n i s q u a d r a t i c

i n x^{2} a n d i t c a n b e s h o w n t h a t

e v e n x^{2} i s n o t i n t e g e r .

W e h a v e n a r r o w e d s e a r c h

t o y t o b e e v e n n u m b e r s .

Answered by Rasheed Soomro last updated on 11/Jan/16

{ax}^{a} + {by}^{b} = {cz}^{c}

L e t a = {4 j}^{2}, b = {4 k}^{2}, c = 4 l^{2}

{4 j}^{2} x^{{4 j}^{2}} + {4 k}^{2} y^{{4 k}^{2}} = 4 l^{2} z^{4 l^{2}}

j^{2} x^{{4 j}^{2}} + k^{2} y^{{4 k}^{2}} = l^{2} z^{4 l^{2}}

{({jx}^{{2 j}^{2}})}^{2} + {({ky}^{{2 k}^{2}})}^{2} = {(l z^{2 l^{2}})}^{2}

({jx}^{{2 j}^{2}}, {ky}^{{2 k}^{2}}, l z^{2 l^{2}}) i s P y t h a g o r e a n t r i p l e .

- - - - - - - - - - -

If (X, Y, Z) i s P y t h a g o r e a n t r i p l e t

s u c h t h a t j ∣ X, k ∣ Y a n d l ∣ Z

(\frac{1}{j} X^{\frac{1}{{2 j}^{2}}}, \frac{1}{k} Y^{\frac{1}{2 k^{2}}}, \frac{1}{l} Z^{\frac{1}{2 l^{2}}}) i s t r i p l e t

s a t i s f y i n g {ax}^{a} + {by}^{b} = {cz}^{c} w h e r e

j = \frac{\sqrt{a}}{2}, k = \frac{\sqrt{b}}{2}, l = \frac{\sqrt{c}}{2}

a, b, c perfect squares and even

- - - - - - - - OR - - - - - - -

If (X, Y, Z) i s P y t h a g o r e a n t r i p l e t

s u c h t h a t \frac{\sqrt{a}}{2} ∣ X, \frac{\sqrt{b}}{2} ∣ Y and \frac{\sqrt{c}}{2} ∣ Z

Also X^{\frac{1}{a}}, Y^{\frac{1}{b}}, Z^{\frac{1}{c}} are integers .

(\frac{2}{\sqrt{a}} X^{\frac{2}{a}}, \frac{2}{\sqrt{b}} Y^{\frac{2}{b}}, \frac{2}{\sqrt{c}} Z^{\frac{2}{c}}) is a required triple .

a, b, c are perfect squares and even .

j = 1, k = 2, l = 3

{(x^{2})}^{2} + {({2 y}^{8})}^{2} = {({3 z}^{18})}^{2}

x^{4} + {4 y}^{16} = {9 z}^{36}

- - - - - -

(m^{2} - n^{2}, 2 m n, m^{2} + n^{2}) i s g e n e r a l

t r i p l e, s u g g e s t e d b y E u c l i d

\forall m, n \in N \land m > n

- - - - - - - - - -

For 2 mn to be of {bx}^{b} type

let m = \frac{b}{2} n^{b - 1}

2 mn = {bn}^{b}

m^{2} + n^{2} = {({bn}^{b})}^{2} + n^{2} = b^{2} n^{2 b} + n^{2}

= (b^{2} n^{2 b - 2} + 1) n^{2} = {cn}^{c}

(b^{2} n^{2 b - 2} + 1) = {cn}^{c - 2}

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