I-have-no-formal-background-in-number-theory-but-I-m-curious-of-how-to-find-positive-integer-solutions-x-y-z-to-the-equation-x-n-y-n-z-n-for-n-Z-Fermat-s-last-theorem-led-me-to-this-Tell

Question Number 4203 by Yozzii last updated on 01/Jan/16

I h a v e n o f o r m a l b a c k g r o u n d i n

n u m b e r t h e o r y, b u t I^{'} m c u r i o u s

o f h o w t o f i n d p o s i t i v e i n t e g e r s o l u t i o n s

(x, y, z) t o t h e e q u a t i o n x^{n} + y^{n} = z^{n} f o r

n \in Z^{-} . {F e r m a t}^{'} s l a s t t h e o r e m l e d

m e t o t h i s . T e l l m e a b o u t t h e c a s e s

o f n = - 1, n = - 2 a n d n = - 3 .

Commented by 123456 last updated on 01/Jan/16

what if n \in Q ?

Commented by Rasheed Soomro last updated on 02/Jan/16

9^{1 / 2} + 16^{1 / 2} = 49^{1 / 2}

64^{- 1 / 2} + 64^{- 1 / 2} = 16^{- 1 / 2}

Answered by RasheedSindhi last updated on 01/Jan/16

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

g e n e r a l s o l u t i o n, n o t i n i n d i v i d u a l

s o l u t i o n s!

I^{'} v e a t t e m p t f o r i n d i v i d u a l o n e s .

x^{- 1} + y^{- 1} = z^{- 1}

\frac{1}{\underset{}{x}} + \frac{1}{y} = \frac{1}{z} \Rightarrow y z + x z = x y

z = \frac{x y}{x + y}

x = y = 8

z = \frac{64}{16} = 4

- - - - - - - -

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

\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{1}{z^{2}}

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

z^{2} = \frac{x^{2} y^{2}}{x^{2} + y^{2}}

x = y = 4, z = 8

Commented by Yozzii last updated on 01/Jan/16

Y e s, g e n e r a l s o l u t i o n s .

Commented by Rasheed Soomro last updated on 01/Jan/16

O n e g e n e r a l s o l u t i o n f o r n = - 1 :

x = y = 2^{m}, z = 2^{m - 1}, \forall m ⩾ 1 \land m \in Z

V e r i f i c a t i o n :

{(2^{m})}^{- 1} + {(2^{m})}^{- 1} = {(2^{m - 1})}^{- 1}

2 . 2^{- m} = 2^{- m + 1}

2^{1 - m} = 2^{1 - m}

Commented by Yozzii last updated on 01/Jan/16

H o w d o y o u p r o v e t h i s a n s w e r ?

I f y o u g u e s s e d i t, w h a t l e d y o u

t o t h i s g u e s s ?

Commented by Rasheed Soomro last updated on 01/Jan/16

x^{- 1} + y^{- 1} = z^{- 1} \Rightarrow z = \frac{x y}{x + y}

z w i l l b e i n t e g e r i n c a s e x + y ∣ x y

A s s u m i n g s p e c i a l c a s e x = y

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

A s z i s i n t e g e r s o 2 ∣ x \Rightarrow x \in E

x = y = 2 k a n d z = k, \forall k \in N (More simpler condition)

V e r i f i c a t i o n :

{(2 k)}^{- 1} + {(2 k)}^{- 1} = k^{- 1}

2 {(2 k)}^{- 1} = k^{- 1}

(2 . 2^{- 1}) . k^{- 1} = k^{-}

k^{- 1} = k^{- 1}

Commented by Yozzii last updated on 01/Jan/16

T h a n k y o u .

Commented by Rasheed Soomro last updated on 01/Jan/16

A s s u m i n g s p e c i a l c a s e

y = 2 x

z = \frac{x y}{x + y} \Rightarrow z = \frac{x . 2 x}{x + 2 x} = \frac{2 x^{2}}{3 x} = \frac{2 x}{3}

z i s i n t e g e r \Rightarrow 3 ∣ x

H e n c e L e t x = 3 k

y = 2 (3 k) = 6 k

z = \frac{2 x}{3} = \frac{2 (3 k)}{3} = 2 k

(x, y, z) = (3 k, 6 k, 2 k)

V e r i f i c a t i o n :

{(3 k)}^{- 1} + {(6 k)}^{- 1} = {(2 k)}^{- 1}

\frac{1}{3 k} + \frac{1}{6 k} = \frac{1}{2 k} \Rightarrow \frac{1}{3} + \frac{1}{6} = \frac{1}{2}

\Rightarrow \frac{2 + 1}{6} = \frac{1}{2}

\Rightarrow \frac{1}{2} = \frac{1}{2}

Commented by Rasheed Soomro last updated on 03/Jan/16

A s s u m i n g y = m x

z = \frac{x y}{x + y} = \frac{x . m x}{x + m x} = \frac{{m x}^{2}}{(1 + m) x} = \frac{m x}{1 + m}

z i s i n t e g e r \Rightarrow 1 + m ∣ x

L e t x = (1 + m) k

y = m (1 + m) k

z = \frac{m (1 + m) k}{1 + m} = m k

(x, y, z) = (m + 1, m (m + 1), m) \forall m \in Z^{+}

Commented by Rasheed Soomro last updated on 03/Jan/16

A s s u m e y = \frac{p}{q} x, \frac{p}{q} i s i n r e d u c e d f o r m

z = \frac{x y}{x + y} = \frac{x (\frac{p}{q} x)}{x + \frac{p}{q} x} = \frac{\frac{p}{q} x^{2}}{\frac{p + q}{q} x} = \frac{p x}{p + q}

z i s i n t e g e r \Rightarrow p + q ∣ x

L e t x = (p + q) k

y = \frac{p (p + q) k}{q}

z = \frac{p}{p + q} \times (p + q) k = p k

N o w c o n s i d e r y = \frac{p (p + q) k}{q}

y i s i n t e g e r \Rightarrow q ∣ p \lor q ∣ p + q \lor q ∣ k

S i n c e (p, q) = 1 s o q ∤ p \land q ∤ p + q

H e n c e q ∣ k

L e t k i s r e p l a c e d b y q k

x = (p + q) q k

y = p (p + q) k

z = p q k

O r

x = q (p + q)

y = p (p + q)

z = p q

W h e r e p, q \in N

Answered by 123456 last updated on 02/Jan/16

i think we can use tha fermat theorem

to that

by fermat last theorem there no other

solution than the trivials one to

x^{n} + y^{n} = z^{n}, n \in N, n > 2

so by this there no other rational solution to

x^{n} + y^{n} = z^{n}

because if it was

x = \frac{p_{x}}{q_{x}}, y = \frac{p_{y}}{q_{y}}, z = \frac{p_{z}}{q_{z}}

x^{n} + y^{n} = z^{n}

\frac{p_{x}^{n}}{q_{x}^{n}} + \frac{p_{y}^{n}}{q_{y}^{n}} = \frac{p_{z}^{n}}{q_{z}^{n}}

{(p_{x} q_{y} q_{z})}^{n} + {(q_{x} p_{y} q_{z})}^{n} = {(q_{x} q_{y} p_{z})}^{n}

a^{n} + b^{n} = c^{n}

and note that

x^{- n} = \frac{1}{x^{n}} = {(\frac{1}{x})}^{n} = a^{n} (rational solution)

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