Solve for +ve integers >0. x^2 +y^4 =z^6


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

$Solve for + ve integers > 0 .$ $x^{2} + y^{4} = z^{6}$
Commented byRasheed 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

	$x^{2} + {(y^{2})}^{2} = {(z^{3})}^{2}$ $(x, y^{2}, z^{3}) i s P y t h a g o r e a n t r i p l e t .$ $A s p e c i a l t y p e P y t h a g o r e a n t r i p l e t$ $w h o s e o n e o f t h e f i r s t t w o m e m b e r s$ $i s p e r f e c t s q u a r e a n d t h i r d m e m b e r$ $i s p e r f e c t c u b e .$ $- - - - - - - - - - - - - - - - - -$ $F o r a l l m, n \in Z^{+} \land m > n$ $(m^{2} - n^{2}, 2 m n, m^{2} + n^{2}) i s a P y t h a g o r e a n$ $t r i p l e t .$ $- - - - - - - - - - - - - - - - - - - - -$ $T r y i n g 2 m n t o b e p e r f e c t s q u a r e$ $I l e t m = 2 n$ $(W e a l s o c a n l e t m = 8 n t h i s w i l l l e d a n$ $o t h e r s o l u t i o n . O r m = 2^{2 k - 1} n)$ $({(2 n)}^{2} - n^{2}, 4 n^{2}, {(2 n)}^{2} + n^{2}) = (3 n^{2}, 4 n^{2}, 5 n^{2})$ $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
	$General Solution x^2 +(y)^2 =(z^3 )^2 (x,y^2 ,z^3 ) is Pythagorean triplet. −−−−−−−−−−−−−−−− For m>n and m,n∈N one general Pythagorean triplet is (m^2 −n^2 ,2mn,m^2 +n^2 ) −−−−−−−−−−−−−−−−− To make 2mn perfect square Let m=2^(2k−1) n 2mn=2(2^(2k−1) n)n=2^(2k) n^2 m^2 +n^2 =(2^(2k−1) n)^2 +n^2 =(2^(4k−2) +1)n^2 In order to make m^2 +n^2 perfect cube Let n=2^(4k−2) +1 ∴ m=2^(2k−1) n=2^(2k−1) ×(2^(4k−2) +1)=2^(6k−3) +2^(2k−1) −−−−−−− m^2 −n^2 =(2^(6k−3) +2^(2k−1) )^2 −(2^(4k−2) +1)^2 =2^(12k−6) +2^(4k−2) +2^(8k−3) −2^(8k−4) −1−2^(4k−1) 2mn=2^(2k) (2^(4k−2) +1)^2 ={2^k (2^(4k−2) +1)}^2 m^2 +n^2 =(2^(4k−2) +1)^3 {(2^(6k−3) +2^(2k−1) )^2 −(2^(4k−2) +1)^2 }^2 +{ 2^k (2^(4k−2) +1)}^4 =(2^(4k−2) +1)^6 ∀ k∈N (x,y,z)=((2^(6k−3) +2^(2k−1) )^2 −(2^(4k−2) +1)^2 , 2^k (2^(4k−2) +1),2^(4k−2) +1) For k=1 75^2 +10^4 =5^6$
	$General Solution$ $x^{2} + {(y)}^{2} = {(z^{3})}^{2}$ $(x, y^{2}, z^{3}) i s P y t h a g o r e a n t r i p l e t .$ $- - - - - - - - - - - - - - - -$ $F o r m > n a n d m, n \in N o n e g e n e r a l$ $P y t h a g o r e a n t r i p l e t i s$ $(m^{2} - n^{2}, 2 m n, m^{2} + n^{2})$ $- - - - - - - - - - - - - - - - -$ $T o m a k e 2 m n p e r f e c t s q u a r e$ $L e t m = 2^{2 k - 1} n$ $2 m n = 2 (2^{2 k - 1} n) n = 2^{2 k} n^{2}$ $m^{2} + n^{2} = {(2^{2 k - 1} n)}^{2} + n^{2} = (2^{4 k - 2} + 1) n^{2}$ $I n o r d e r t o m a k e m^{2} + n^{2} p e r f e c t c u b e$ $L e t n = 2^{4 k - 2} + 1$ $∴ m = 2^{2 k - 1} n = 2^{2 k - 1} \times (2^{4 k - 2} + 1) = 2^{6 k - 3} + 2^{2 k - 1}$ $- - - - - - -$ $m^{2} - n^{2} = {(2^{6 k - 3} + 2^{2 k - 1})}^{2} - {(2^{4 k - 2} + 1)}^{2}$ $= 2^{12 k - 6} + 2^{4 k - 2} + 2^{8 k - 3} - 2^{8 k - 4} - 1 - 2^{4 k - 1}$ $2 m n = 2^{2 k} {(2^{4 k - 2} + 1)}^{2} = {2^{k} (2^{4 k - 2} + 1)}^{2}$ $m^{2} + n^{2} = {(2^{4 k - 2} + 1)}^{3}$ ${{(2^{6 k - 3} + 2^{2 k - 1})}^{2} - {(2^{4 k - 2} + 1)}^{2}}^{2} + {2^{k} (2^{4 k - 2} + 1)}^{4}$ $= {(2^{4 k - 2} + 1)}^{6} \forall k \in N$ $(x, y, z) = ({(2^{6 k - 3} + 2^{2 k - 1})}^{2} - {(2^{4 k - 2} + 1)}^{2}, 2^{k} (2^{4 k - 2} + 1), 2^{4 k - 2} + 1)$ $F o r k = 1 75^{2} + 10^{4} = 5^{6}$
	Commented byprakash 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 byRasheed Soomro last updated on 06/Jan/16

	$G^{OO} D T echnique! T H α n k S!$
	Commented byRasheed 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 byRasheed Soomro last updated on 07/Jan/16

	$(x, y, z) i s P y t h a g o r e a n t r i p l e t$ $\Rightarrow ({x y}^{3} z^{2}, y^{2} z, z y) i s r e q u i r e d t r i p l e t .$ $Let x, y, z are m^{2} - n^{2}, 2 mn, m^{2} + n^{2} type .$ ${xy}^{3} z^{2} = (m^{2} - n^{2}) {(2 mn)}^{3} {(m^{2} + n^{2})}^{2}$ $y^{2} z = {(2 mn)}^{2} (m^{2} + n^{2})$ $yz = (2 mn) (m^{2} + n^{2})$ $General Triplet for x^{2} + y^{4} = z^{6} :$ $(\frac{(m^{2} - n^{2}) {(2 mn)}^{3} {(m^{2} + n^{2})}^{2},}{} \frac{{(2 mn)}^{2} (m^{2} + n^{2}),}{} \frac{(2 mn) (m^{2} + n^{2})}{})$ $Or$ $(\frac{{(m^{2} - n^{2})}^{3} (2 mn) {(m^{2} + n^{2})}^{2},}{} \frac{{(m^{2} - n^{2})}^{2} (m^{2} + n^{2}),}{} \frac{(m^{2} - n^{2}) (m^{2} + n^{2})}{})$ $\forall m, n \in N w i t h m > n$
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