Solve-for-ve-integers-gt-0-x-2-y-4-z-6-

Question Number 4230 by prakash jain last updated on 03/Jan/16

Solve for + ve integers > 0 .

x^{2} + y^{4} = z^{6}

Commented by Rasheed Soomro last updated on 04/Jan/16

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

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

w h i c h f i r s t / s e c o n d m e m b e r i s p e r f e c t s q u a r e

a n d t h i r d m e m b e r i s p e r f e c t c u b e .

75, 100, 125

75^{2} + 100^{2} = 125^{2}

75^{2} + 10^{4} = 5^{6}

Answered by Rasheed Soomro last updated on 05/Jan/16

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

(x, y^{2}, z^{3}) i s P y t h a g o r e a n t r i p l e t .

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

w h o s e o n e o f t h e f i r s t t w o m e m b e r s

i s p e r f e c t s q u a r e a n d t h i r d m e m b e r

i s p e r f e c t c u b e .

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F o r a l l m, n \in Z^{+} \land m > n

(m^{2} - n^{2}, 2 m n, m^{2} + n^{2}) i s a P y t h a g o r e a n

t r i p l e t .

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T r y i n g 2 m n t o b e p e r f e c t s q u a r e

I l e t m = 2 n

(W e a l s o c a n l e t m = 8 n t h i s w i l l l e d a n

o t h e r s o l u t i o n . O r m = 2^{2 k - 1} n)

({(2 n)}^{2} - n^{2}, 4 n^{2}, {(2 n)}^{2} + n^{2}) = (3 n^{2}, 4 n^{2}, 5 n^{2})

T r y i n g 5 n^{2} t o b e p e r f e c t c u b e I g u e s s e d n = 5 *

m = 2 n = 2 (5) = 10

S o (m^{2} - n^{2}, 2 m n, m^{2} + n^{2})

= (10^{2} - 5^{2}, 2 (10) (5), 10^{2} + 5^{2})

= (75, 100, 125)

∴ 75^{2} + 100^{2} = 125^{2}

O r 75^{2} + {(10^{2})}^{2} = {(5^{3})}^{2}

75^{2} + 10^{4} = 5^{6}

* A n o t h e r g u e s s f o r 5 n^{2} t o b e p e r f e c t c u b e

5 n^{2} = 5 . 5^{2} . 4^{3} \Rightarrow n = 5 . 2^{3} = 40

m = 2 n = 2 (40) = 80

(m^{2} - n^{2}, 2 m n, m^{2} + n^{2})

= (80^{2} - 40^{2}, 2 (80) (40), 80^{2} + 40^{2})

= (6400 - 1600, 6400, 6400 + 1600)

= (4800, 6400, 8000)

∴ 4800^{2} + 6400^{2} = 8000^{2}

4800^{2} + {(80^{2})}^{2} = {(20^{3})}^{2}

4800^{2} + 80^{4} = 20^{6}

Answered by Rasheed Soomro last updated on 05/Jan/16

General Solution

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

(x, y^{2}, z^{3}) i s P y t h a g o r e a n t r i p l e t .

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F o r m > n a n d m, n \in N o n e g e n e r a l

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

(m^{2} - n^{2}, 2 m n, m^{2} + n^{2})

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T o m a k e 2 m n p e r f e c t s q u a r e

L e t m = 2^{2 k - 1} n

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

m^{2} + n^{2} = {(2^{2 k - 1} n)}^{2} + n^{2} = (2^{4 k - 2} + 1) n^{2}

I n o r d e r t o m a k e m^{2} + n^{2} p e r f e c t c u b e

L e t n = 2^{4 k - 2} + 1

∴ m = 2^{2 k - 1} n = 2^{2 k - 1} \times (2^{4 k - 2} + 1) = 2^{6 k - 3} + 2^{2 k - 1}

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m^{2} - n^{2} = {(2^{6 k - 3} + 2^{2 k - 1})}^{2} - {(2^{4 k - 2} + 1)}^{2}

= 2^{12 k - 6} + 2^{4 k - 2} + 2^{8 k - 3} - 2^{8 k - 4} - 1 - 2^{4 k - 1}

2 m n = 2^{2 k} {(2^{4 k - 2} + 1)}^{2} = {2^{k} (2^{4 k - 2} + 1)}^{2}

m^{2} + n^{2} = {(2^{4 k - 2} + 1)}^{3}

{{(2^{6 k - 3} + 2^{2 k - 1})}^{2} - {(2^{4 k - 2} + 1)}^{2}}^{2} + {2^{k} (2^{4 k - 2} + 1)}^{4}

= {(2^{4 k - 2} + 1)}^{6} \forall k \in N

(x, y, z) = ({(2^{6 k - 3} + 2^{2 k - 1})}^{2} - {(2^{4 k - 2} + 1)}^{2}, 2^{k} (2^{4 k - 2} + 1), 2^{4 k - 2} + 1)

F o r k = 1 75^{2} + 10^{4} = 5^{6}

Commented by prakash jain last updated on 05/Jan/16

A good formula .

An observation for specific case :

75^{2} + 10^{4} = 5^{6}

5^{4} \cdot 3^{2} + 5^{4} . 2^{4} = 5^{4} \cdot 5^{2}

So if x^{2} + y^{2} = z^{2} and y is a perfect square say u^{2}

then we can get solution for x^{2} + y^{4} = z^{6}

by multiplying x, y, z by z^{4} .

Commented by Rasheed Soomro last updated on 06/Jan/16

G^{OO} D T echnique! T H α n k S!

Commented by Rasheed Soomro last updated on 08/Jan/16

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

▽

T r i p l e s a t i s f y i n g x^{2} + y^{4} = z^{6}

Your technique applying twice

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

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

{(x y)}^{2} + y^{4} = {(z y)}^{2}

{(x y)}^{2} {(z y)}^{4} + y^{4} {(z y)}^{4} = {(z y)}^{2} {(z y)}^{4}

{({x y}^{3} z^{2})}^{2} + {(y^{2} z)}^{4} = {(z y)}^{6}

(x, y, z) i s P y t h a g o r e a n t r i p l e t

\Rightarrow ({x y}^{3} z^{2}, y^{2} z, z y) i s t r i p l e t f o r x^{2} + y^{4} = z^{6} .

(3, 4, 5) \Rightarrow (3 {(4)}^{3} {(5)}^{2}, {(4)}^{2} (5), (5) (4))

= (3 (64) (25), 16 (5), 20) = (4800, 80, 20)

(4, 3, 5) \Rightarrow (4 {(3)}^{3} {(5)}^{2}, {(3)}^{2} (5), (3) (5)) = (2700, 45, 15)

Commented by Rasheed Soomro last updated on 07/Jan/16

(x, y, z) i s P y t h a g o r e a n t r i p l e t

\Rightarrow ({x y}^{3} z^{2}, y^{2} z, z y) i s r e q u i r e d t r i p l e t .

Let x, y, z are m^{2} - n^{2}, 2 mn, m^{2} + n^{2} type .

{xy}^{3} z^{2} = (m^{2} - n^{2}) {(2 mn)}^{3} {(m^{2} + n^{2})}^{2}

y^{2} z = {(2 mn)}^{2} (m^{2} + n^{2})

yz = (2 mn) (m^{2} + n^{2})

General Triplet for x^{2} + y^{4} = z^{6} :

(\frac{(m^{2} - n^{2}) {(2 mn)}^{3} {(m^{2} + n^{2})}^{2},}{} \frac{{(2 mn)}^{2} (m^{2} + n^{2}),}{} \frac{(2 mn) (m^{2} + n^{2})}{})

Or

(\frac{{(m^{2} - n^{2})}^{3} (2 mn) {(m^{2} + n^{2})}^{2},}{} \frac{{(m^{2} - n^{2})}^{2} (m^{2} + n^{2}),}{} \frac{(m^{2} - n^{2}) (m^{2} + n^{2})}{})

\forall m, n \in N w i t h m > n