x,y,z ∈ Z^+ find all solutions of xy=(x+y)z


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Question Number 69939 by mr W last updated on 29/Sep/19

$x, y, z \in Z^{+}$ $f i n d a l l s o l u t i o n s o f x y = (x + y) z$
Commented by Rasheed.Sindhi last updated on 29/Sep/19

$k \in Z^{+} i n t h e f o l l o w i n g .$ $(x, y, z) = (2 k, 2 k, k),$ $(5 k, 20 k, 4 k), (20 k, 5 k, 4 k),$ $(3 k, 6 k, 2 k), (6 k, 3 k, 2 k), (7 k, 42 k, 6 k),$ $(4, 12, 3)$ $^{∙} I f (p, q, r) i s a s o l u t i o n t h e n (q, p, r)$ $i s a l s o s o l u t i o n .$ $^{∙} I f (p, q, r) i s a s o l u t i o n t h e n (p k, q k, r k)$ $i s a l s o s o l u t i o n . k \in Z^{+}$ $. . . .$
Commented by mr W last updated on 29/Sep/19

$t h a n k s s i r!$ $t h e s e a r e s o l u t i o n s . b u t h o w c a n w e$ $e x p r e s s A L L p o s s i b l e s o l u t i o n s ?$
Commented by Rasheed.Sindhi last updated on 29/Sep/19

$S i r a t t h e m o m e n t I h a v e n o i d e a o f$ $e x p r e s s i n g a l l t h e p o s s i b l e s o l u t i o n s!$ $H o w e v e r I a m t h i n k i n g a b o u t i t . . .$
Commented by mr W last updated on 29/Sep/19

$t h a n k y o u f o r t r y i n g s i r!$
Commented by Rasheed.Sindhi last updated on 29/Sep/19

$S i r I t h i n k i n f i n i t e s o l u t i o n s!$ $I^{'} l l t r y t o w r i t e s o m e o b s e r v a t i o n s$ $a b o u t s u c h t r i p l e t s w h i c h s a t i s f y$ $y o u r e q u a t i o n .$
Commented by mind is power last updated on 29/Sep/19
$suppose x and y coprime ⇒x+y and xy are coprim ⇒xy∣z⇒z=xy and x+y=1 now assume x and y are note coprim let d=gcd(x,y)⇒x=dx′,y=dy′ x′,y′ coprime ⇒d^2 x′y′=d(x′+y′)z ⇒dx′y′=(x′+y′)z x′y′∣z cause x′+y′ and x′y′ are coprime and x′y′∣dx′y′=(x′+y′)z z=x′y′.k ⇒d=k(x′+y′) let S={n∈N n∣d} for all m∈N^∗ we can Whrite m=p+q withe p,q coprim m=1+(m−1) k=m withe m∈S x′+y′=(d/m)∈S (x′,y′)∈{(a,b)∣a+b=(d/m),and gcd(a,b)=1} x=dx′=da y=dx′=db z=kx′y′=mab$
$s u p p o s e x a n d y c o p r i m e$ $\Rightarrow x + y a n d x y a r e c o p r i m$ $\Rightarrow x y ∣ z \Rightarrow z = x y a n d x + y = 1$ $n o w a s s u m e x a n d y a r e n o t e c o p r i m$ $l e t d = g c d (x, y) \Rightarrow x = {d x}^{'}, y = {d y}^{'}$ $x^{'}, y^{'} c o p r i m e$ $\Rightarrow d^{2} x^{'} y^{'} = d (x^{'} + y^{'}) z$ $\Rightarrow {d x}^{'} y^{'} = (x^{'} + y^{'}) z$ $x^{'} y^{'} ∣ z c a u s e x^{'} + y^{'} a n d x^{'} y^{'} a r e c o p r i m e$ $a n d x^{'} y^{'} ∣ {d x}^{'} y^{'} = (x^{'} + y^{'}) z$ $z = x^{'} y^{'} . k$ $\Rightarrow d = k (x^{'} + y^{'})$ $l e t S = {n \in N n ∣ d}$ $f o r a l l m \in N^{*} w e c a n W h r i t e m = p + q w i t h e p, q c o p r i m m = 1 + (m - 1)$ $k = m w i t h e m \in S$ $x^{'} + y^{'} = \frac{d}{m} \in S$ $(x^{'}, y^{'}) \in {(a, b) ∣ a + b = \frac{d}{m}, a n d g c d (a, b) = 1}$ $x = {d x}^{'} = d a$ $y = {d x}^{'} = d b$ $z = {k x}^{'} y^{'} = m a b$
Commented by Prithwish sen last updated on 30/Sep/19
$Given xy=xz+yz ⇒ 2xy−2xz−2yz = 0 ......(i) and 2xy+2xz+2yz=(x+y+z)^2 −(x^2 +y^2 +z^2 ).....(ii) Adding (i) and (ii) ⇒xy=(((x+y+z)^2 −(x^2 +y^2 +z^2 ))/4) .....(iii) Now there are three possibilities x y z Case I even odd even Case II even even odd CaseIII even even even Now here I am considering only Case I Let x= 2m y= 2n+1 z = 2p ⇒ p = ((m(2n+1))/(2(m+n)+1)) ⇒ 2mp+2np+p = 2mn +m ........(iv) Now from (iii) xy = (((2m+2n+2p+1)^2 −{4m^2 +(2n+1)^2 +4p^2 })/4) = ((8mn+8mp+8np+4m+4p)/4) =( 2mp+2np+p)+2mn+m = 4mn +2m Now ∵ p is an integer ∴ p = n+ ((m−2n^2 −n)/(2m+2n+1)) ⇒m−2n^2 −n=0 ⇒ n = (((√(1+8m))−1)/4) = (((1+4q)−1)/4) = q putting m = 2q^2 +q p =(((2q^2 +q)(2q+1))/(2{(2q^2 +q)+q}+1)) = ((q(2q+1)^2 )/(2q(2q+1)+2q+1)) =q ∴ x = 2q(2q+1) , y =2q+1 and z = 2q q x y z 1 6 3 2 2 20 5 4 3 42 7 6 4 72 9 8 and so on ......... m,n,p ,q are integers Mr. W Sir and Rasheed Sir please give your valuable suggestions.Sir I make the whole thing only dependent on q. By putting q=1,2,... and you get the result.$
$Given$ $xy = xz + yz$ $\Rightarrow 2 xy - 2 xz - 2 yz = 0 . . . . . . (i)$ $and 2 xy + 2 xz + 2 yz = {(x + y + z)}^{2} - (x^{2} + y^{2} + z^{2}) . . . . . (ii)$ $Adding (i) and (ii)$ $\Rightarrow xy = \frac{{(x + y + z)}^{2} - (x^{2} + y^{2} + z^{2})}{4} . . . . . (iii)$ $Now there are three possibilities$ $x y z$ $Case I even odd even$ $Case I I even even odd$ $CaseIII even even even$ $Now here I am considering only Case I$ $Let x = 2 m y = 2 n + 1 z = 2 p$ $\Rightarrow p = \frac{m (2 n + 1)}{2 (m + n) + 1}$ $\Rightarrow 2 mp + 2 np + p = 2 mn + m . . . . . . . . (iv)$ $Now from (iii)$ $xy = \frac{{(2 m + 2 n + 2 p + 1)}^{2} - {4 m^{2} + {(2 n + 1)}^{2} + 4 p^{2}}}{4}$ $= \frac{8 mn + 8 mp + 8 np + 4 m + 4 p}{4}$ $= (2 mp + 2 np + p) + 2 mn + m = 4 mn + 2 m$ $Now ∵ p is an integer$ $∴ p = n + \frac{m - {2 n}^{2} - n}{2 m + 2 n + 1}$ $\Rightarrow m - {2 n}^{2} - n = 0$ $\Rightarrow n = \frac{\sqrt{1 + 8 m} - 1}{4} = \frac{(1 + 4 q) - 1}{4} = q putting m = 2 q^{2} + q$ $p = \frac{(2 q^{2} + q) (2 q + 1)}{2 {(2 q^{2} + q) + q} + 1} = \frac{q {(2 q + 1)}^{2}}{2 q (2 q + 1) + 2 q + 1}$ $= q$ $∴ x = 2 q (2 q + 1), y = 2 q + 1 and z = 2 q$ $q x y z$ $1 6 3 2$ $2 20 5 4$ $3 42 7 6$ $4 72 9 8$ $and so on . . . . . . . . .$ $m, n, p, q are integers$ $Mr . W Sir and Rasheed Sir please give your$ $valuable suggestions . Sir I make the whole thing$ $only dependent on q . By putting q = 1, 2, . . .$ $and you get the result .$
Commented by mr W last updated on 29/Sep/19

$s i r m i n d i s p o w e r :$ $t h a n k s f o r a t t e m p t i n g s i r!$ $b u t$ $x = d a$ $y = d b$ $z = {k d}^{2} a b$ ${d o n}^{'} t s a t i s f y x y = (x + y) z . b e s i d e s$ $h o w t o g e t a, b, d, k ?$
Commented by mind is power last updated on 29/Sep/19

$i f i x i s w i t c h z a n d k$
Commented by mr W last updated on 29/Sep/19

$p r i t h w i s h s i r :$ $t h a n k s f o r t r y i n g!$ $w i t h x = 2 m, y = \frac{\sqrt{1 + 8 m} + 1}{2}, z = 2 p$ $i f y o u t a k e m = 2, y \neq i n t e g e r!$
Commented by mr W last updated on 29/Sep/19

$t h e r e a r e t h r e e u n k n o w n s x, y, z .$ $a g e n e r a l s o l u t i o n m u s t h a v e t w o$ $p a r a m e t e r s i n t h e e x p r e s s i o n . t h i s$ $i s l i k e t o f i n d t h e g e n e r a l i n t e g e r$ $s o l u t i o n f o r x + 2 y + z = 7 .$
Commented by mind is power last updated on 29/Sep/19

$d g i v e n p a r a m e t e r$ $z = d a b$ $m d i v u s o r o f d$ $a a n d b a r e s u c h a + b = \frac{d}{m}$ $a n d g c d (a, b) = 1$ $a = \frac{d}{m} - b$ $g c d (a, b) = 1 \Rightarrow g c d (\frac{d}{m}, b)$ $s o i t b i s c o p r i m e w i t h e \frac{d}{m} w i t h e b ⩽ \frac{d}{m}$ $s a m e w i t h e a c o p r i m e w i t h e \frac{d}{m}$ $(a, b) * d e s c r i b e a l l i n t e g e r w i t h e a + b = \frac{d}{m} w i t h e a a n d b c o p r i m e w i t h e \frac{d}{m}$ $e x a m p l a + b = 6$ $(a, b) \in {(1, 5) (5, 1)}$ $S o r r y m y e n g l i s h n o t G o o d$
	Answered by Rasheed.Sindhi last updated on 02/Oct/19

	$I h a v e m a d e a f o r m u l a w h i c h p r o d u c e$ $a t r i p l e t s a t i s f y i n g t h e g i v e n e q u a t i o n$ $f o r e v e r y p a i r o f p o s i t i v e i n t e g e r s :$ $L e t p, q \in Z^{+}$ $x = p (p + q)$ $y = q (p + q)$ $z = p q$ $W e c a n a l g e b r a i c a l l y p r o v e t h a t$ $(x, y, z) = (p (p + q), q (p + q), p q) i s s o l u t i o n$ $o f x y = (x + y) z$ $T h i s p r o v e s t h a t t h e s o l u t i o n s a r e$ $i n f i n i t e .$ $C l a i m :$ $^{∙} I f p & q b e c o p r i m e t h e n$ $(x, y, z) w i l l b e p r i m i t i v e t r i p l e s a t i s f y i n g$ $x y = (x + y) z$ $^{∙} T h e a b o v e f o r m u l a s c a n p r o d u c e$ $a l l t h e p o s s i b l e p r i m i t i v e s o l u t i o n s$ $^{∙} N o n - p r i m i t i v e s o l u t i o n a r e m e r e l y$ $i n t e g r a l m u l t i p l e s o f p r i m i t i v e s o l u t i o n s .$ $Y o u r o f f e r e d e x a m p l e ((8, 24, 6)) i s$ $i s n o n - p r i m i t i v e s o l u t i o n a n d i t s$ $c o r r e s p o n d i n g p r i m i t i v e s o l u t i o n$ $i s (4, 12, 3) a n d t h i s c a n b e p r o d u c e d$ $b y m y f o r m u l a s (p u t p = 1, q = 3) .$ $H o w d o y o u s e e t h i s c l a i m . p l e a s e$ $g i v e a c o u n t e r e x a m p l e i f y o u {d o n}^{'} t$ $a g r e e .$
	Commented by Prithwish sen last updated on 30/Sep/19

	$As a whole$ $for the expression$ $xy = (x + y) z x, y, z \in Z^{+}$ $we have$ $Case I x = 2 q (2 q + 1) y = (2 q + 1) z = 2 q$ $Case II x = 2 q y = 2 q (2 q - 1) z = 2 q - 1$ $and for Case III no such expression found .$ $x y z$ $q = 1 6 3 2 - Case I$ $2 2 1 - Case II$ $q = 2 20 5 4 - Case I$ $4 12 3 - Case II (value of$ $q = 3 42 7 6 - Case I x and y$ $6 30 5 - Case II can be$ $q = 4 72 9 8 - Case I altered)$ $8 56 7 - Case II$ $and so on . . . . . .$ $It confirms that the expression has infinite$ $number of solutions .$
	Commented by Rasheed.Sindhi last updated on 03/Oct/19

	$S i r h o w c a n w e p r o v e t h a t t h e a b o v e$ $f o r m u l a s c a n p r o d u c e a l l p o s s i b l e$ $s o l u t i o n s ?$ $&$ ${W h a t}^{'} s t h e s o u r s e o f p r o b l e m ?$
	Commented by mr W last updated on 29/Sep/19

	${t h a t}^{'} s a n i c e w a y s i r!$ $f r o m g i v e n p a n d q w e c a n c o n s t r u c t$ $a s o l u t i o n . b u t t h e r e c o u l d b e m o r e$ $t h a n j u s t o n e s o l u t i o n f o r g i v e n p, q .$ $e . g . z = 6$ $(x, y) = (10, 15) i s a s o l u t i o n, b u t$ $(8, 24) i s a l s o a s o l u t i o n .$
	Commented by Rasheed.Sindhi last updated on 30/Sep/19

	$Y e s s i r, t h e f o r m u l a {d o e s n}^{'} t c o v e r a l l$ $p o s s i b l e s o l u t i o n s . H o w e v e r i t a s s u r e s$ $t h a t t h e r e a r e i n f i n i t e s o l u t i o n s .$ $F o r c o m p o s i t e z, t h e f o r m u l a c a n$ $g i v e m u l t i p l e s o l u t i o n s . F o r e x a m p l e$ $z = 6 \Rightarrow p q = 6 \Rightarrow (p, q) = (1, 6) o r (2, 3)$ $\Rightarrow (x, \bar{y}, z) = (7, 42, 6) o r (10, 15, 6) .$ $B u t a f t e r a l l t h e f o r m u l a {d o e s n}^{'} t$ $c o v e r a l l p o s s i b l e s o l u t i o n s .$
	Commented by Prithwish sen last updated on 30/Sep/19

	$Mr W . Sir please see my solution I have$ $fix my error . I think I get a general formula$ $with a single parameter q .$
	Commented by Prithwish sen last updated on 30/Sep/19

	$Sir please waiting for your response .$
	Commented by mr W last updated on 30/Sep/19

	${i t}^{'} s a s o l u t i o n . b u t i t h i n k i t {d o e s n}^{'} t$ $c o v e r a l l p o s s i b l e s o l u t i o n s . z = 2 q$ $m e a n s z i s a l w a y s e v e n . t h i s i s n o t$ $t r u e, e . g . x = 4, y = 12, z = 3 .$
	Commented by Rasheed.Sindhi last updated on 30/Sep/19

	$P r i t h w i s h s e n s i r$ $Y o u r^{'} p r o d u c t {m a c h i n e}^{'} i s w o n d e r f u l!$ $I t p r o d u c e s s o l u t i o n s s u c c e s s f u l l y .$ $B u t t h e r e a r e s o m e s o l u t i o n s w h i c h$ ${c a n}^{'} t b e p r o d u c e d b y y o u r m a c h i n e .$ $W h e r e a s m r W s i r d e m a n d s s u c h$ $m a c h i n e w h i c h c a n p r o d u c e a l l p o s s i b l e$ $s o l u t i o n s!$ $F o r e x a m p l e f o r z = 6 y o u r m a c h i n e$ $(f a r m u l a) p r o d u c e s s i n g l e t r i p l e t$ $(s o l u t i o n) : (42, 7, 6) b y p u t t i n g q = 3 .$ $W h e r e a s m a n y o t h e r t r i p l e t s (w h i c h$ ${a r e n}^{'} t c o m e f r o m y o u r m a c h i n e)$ $s a t i s f y t h e e q u a t i o n x y = (x + y) z .$ $S u c h a s (10, 15, 6) e t c (C a n t h i s b e$ $p r o d u c e d b y y o u r f o r m u l a ?)$
	Commented by Prithwish sen last updated on 30/Sep/19

	$Thank you sir .$
	Commented by mr W last updated on 30/Sep/19

	$t h a n k y o u a l l s i r s!$
	Commented by Prithwish sen last updated on 01/Oct/19

	$Thank you sir .$
	Commented by Rasheed.Sindhi last updated on 01/Oct/19

	$A n E q u i v a l e n t E x p r e s s i o n o f$ $x y = (x + y) z .$ $x y = (x + y) z \Rightarrow x y - y z - z x = 0$ $\Rightarrow x y + y (- z) + (- z) x = 0$ $N o w,$ ${(x + y - z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y (- z) + (- z) x)$ $= x^{2} + y^{2} + z^{2} + 2 (0)$ $x^{2} + y^{2} + z^{2} = {(x + y - z)}^{2}$ $T h i s i s e q u i v a l e n t t o x y = (x + y) z$ $i - e s o l u t i o n s o f b o t h a r e s a m e .$ $N o t e : T h e i d e a r a i s e d i n m y m i n d a f t e r$ $s e e i n g a c o m m e n t o f P r i t h w i s h s e n s i r .$
	Commented by Rasheed.Sindhi last updated on 03/Oct/19

	$m r W s i r$ $B y p r i m i t i v e t r i p l e I m e a n t a t r i p l e$ $w i t h g c d 1$ $(9, 18, 6) = 3 (3, 6, 2)$ $(3, 6, 2) i s p r i m i t i v e t r i p l e .$ $z = 2 \Rightarrow p q = 2 \Rightarrow {p, q} = {1, 2}$ $x = 1 (1 + 2) = 3$ $y = 2 (1 + 2) = 6$ $z = 2 \times 1 = 2$
	Commented by mr W last updated on 03/Oct/19

	$i c r e a t e d t h i s q u e s t i o n b e c a u s e i b e c a m e$ $u n s u r e t o m y s o l u t i o n t o Q 10944 .$
	Commented by Rasheed.Sindhi last updated on 02/Oct/19

	$P r i t h w i s h s e n s i r & m i n d - i s - p o w e r s i r!$ $P l e a s e r e v i e w m y a b o v e a n s w e r a n d$ $g i v e f e e d b a c k .$
	Commented by Rasheed.Sindhi last updated on 02/Oct/19

	$S i r m r W$ $P l e a s e r e v i e w m y a b o v e a n s w e r . I^{'} v e$ $a d d e d s o m e m a t e r i a l i n r e d l i n e s .$ $I^{'} m a n x i o u s l y w a i t i n g f o r y o u r r e s p o n s e .$
	Commented by mr W last updated on 02/Oct/19

	$t h a n k s f o r a l l t h e e f f o r t s s i r!$ $y o u c a n g e t (8, 24, 6) f r o m (4, 12, 3), b u t$ $h o w c a n w e g e t (9, 18, 6) f r o m (4, 12, 3) ?$
	Commented by mr W last updated on 03/Oct/19

	$n o w i s e e . t h a n k s s i r!$ $t h e t a s k i s n o w h o w t o e x p r e s s x a n d y$ $x = f (p, q)$ $y = g (p, q)$ $z = p q$ $s u c h t h a t x a n d y c o v e r a l l p o s s i b i l i t i e s .$ $c a n w e f i n d t h e f (p, q) a n d g (p, q) ?$
	Commented by Rasheed.Sindhi last updated on 03/Oct/19

	$x = f (p, q) = p (p + q)$ $y = g (p, q) = q (p + q)$ $z = h (p, q) = p q$ $F o r a g i v e n z = 12 (s a y)$ $z = 12 \Rightarrow p q = 12 (f o r p r i m i t i v e s o l u t i o n s$ $t a k e p & q c o p r i m e)$ $(p, q) = (1, 12), (3, 4), (4, 3), (12, 1)$ $(1, 12) : x = 1 (1 + 12) = 13$ $y = 12 (1 + 12) = 156$ $z = 12$ $A l l p o s s i b l e p r i m i t i v e s o l u t i o n s w i t h$ $z = 12 a r e : (13, 156, 12), (21, 28, 12),$ $(28, 21, 12) & (156, 13, 12)$ $F o r p r i m i t i v e s o l u t i o n s w i t h z = 12$ $x & y {c a n}^{'} t t a k e o t h e r v a l u e s$ $(I f m y c l a i m i s r i g h t) . I f x & y h a v e$ $v a l u e s o t h e r t h a n a b o v e, t h e n t h e$ $t h a t s o l u t i o n m u s t b e n o n - p r i m i t i v e .$ $p l e a s e t r y c o u n t e r e x a m p l e : (a, b, 12)$ $s u c h t h a t g c d (a, b, 12) = 1 a n d$ $a b = 12 (a + b)$
	Commented by mr W last updated on 03/Oct/19

	$t h a n k s a g a i n s i r!$ $m y q u e s t i o n i s, f o r a n y g i v e n p a n d q,$ $c a n w e f i n d x = f (p, q) a n d y = g (p, q) w h i c h$ $p r o d u c e a l l s o l u t i o n s, p r i m i t i v e a n d$ $n o n - p r i m i t i v e o n e s ?$
	Commented by Rasheed.Sindhi last updated on 03/Oct/19

	$B y a d d i n g o n e a d d i t i o n a l p a r a m e t e r k :$ $x = k p (p + q)$ $y = k q (p + q)$ $z = k p q$ $p & q b e i n g c o p r i m e a n d k a n y p o s i t i v e$ $i n t e g e r . I t h i n k s i r t h e s e f o r m u l a s$ $c a n c o v e r a l l p r i m i t i v e a s w e l l a s$ $n o n - p r i m i t i v e s o l u t i o n s .$
	Commented by mr W last updated on 03/Oct/19

	$b i g b i g t h a n k s s i r!$
	Answered by Rasheed.Sindhi last updated on 02/Oct/19
	$In one of my comments here I have shown that x^2 +y^2 +z^2 =(x+y−z)^2 is equivalent to xy=(x+y)z.I mean both have same solutions. Can this derived form helps us towarxs the goal(Expressing all possible solutions). Observe this new form,it resembles a^2 +b^2 +c^2 =d^2 .And for this there are already established formulas: a∈O; m,n,p,q∈{0,1,2,...}; (m,n,p,q)=1; m+n+p+q)∈O For primitive quadruples (a,b,c,d): a=m^2 +n^2 −p^2 −q^2 b=2(mq+np) c=2(nq−mp), d=m^2 +n^2 +p^2 +q^2 Here d=x+y−z=m^2 +n^2 +p^2 +q^2 Where {x,y,z}={a,b,c } in any order. z=a=m^2 +n^2 −p^2 −q^2 (z is odd) d=(2mq+2np)+(2nq−2mp)−(m^2 +n^2 −p^2 −q^2 ) =m^2 +n^2 +p^2 +q^2 mq+np+nq−mp−m^2 −n^2 =0 q(m+n)+p(n−m)−m^2 −n^2 =0 q=((p(m−n)+m^2 +n^2 )/(m+n)) Now x=b=2m(((p(m−n)+m^2 +n^2 )/(m+n)))+2np y=c=2n(((p(m−n)+m^2 +n^2 )/(m+n)))−2mp z=a=m^2 +n^2 −p^2 −(((p(m−n)+m^2 +n^2 )/(m+n)))^2 (x,y,z) is primitive triple Similarly z=b=2mq+2np (z∈E) .... ... z=c=2nq−2mp(z∈E) ..... These formulas or not of practical use.Because it′s very difficult to choose such m,n & p that q is integer and (m,n,p,q)=1 & other conditions fulfill. Imractical farmulas$
	$I n o n e o f m y c o m m e n t s h e r e I h a v e$ $s h o w n t h a t$ $x^{2} + y^{2} + z^{2} = {(x + y - z)}^{2} i s e q u i v a l e n t$ $t o x y = (x + y) z . I m e a n b o t h h a v e$ $s a m e s o l u t i o n s .$ $C a n t h i s d e r i v e d f o r m h e l p s u s t o w a r x s$ $t h e g o a l (E x p r e s s i n g a l l p o s s i b l e s o l u t i o n s) .$ $O b s e r v e t h i s n e w f o r m, i t r e s e m b l e s$ $a^{2} + b^{2} + c^{2} = d^{2} . A n d f o r t h i s t h e r e a r e$ $a l r e a d y e s t a b l i s h e d f o r m u l a s :$ $a \in O; m, n, p, q \in {0, 1, 2, . . .};$ $(m, n, p, q) = 1; m + n + p + q) \in O$ $F o r p r i m i t i v e q u a d r u p l e s$ $(a, b, c, d) :$ $a = m^{2} + n^{2} - p^{2} - q^{2}$ $b = 2 (m q + n p)$ $c = 2 (n q - m p),$ $d = m^{2} + n^{2} + p^{2} + q^{2}$ $H e r e d = x + y - z = m^{2} + n^{2} + p^{2} + q^{2}$ $W h e r e {x, y, z} = {a, b, c} i n a n y o r d e r .$ $z = a = m^{2} + n^{2} - p^{2} - q^{2} (z i s o d d)$ $d = (2 m q + 2 n p) + (2 n q - 2 m p) - (m^{2} + n^{2} - p^{2} - q^{2})$ $= m^{2} + n^{2} + p^{2} + q^{2}$ $m q + n p + n q - m p - m^{2} - n^{2} = 0$ $q (m + n) + p (n - m) - m^{2} - n^{2} = 0$ $q = \frac{p (m - n) + m^{2} + n^{2}}{m + n}$ $N o w$ $x = b = 2 m (\frac{p (m - n) + m^{2} + n^{2}}{m + n}) + 2 n p$ $y = c = 2 n (\frac{p (m - n) + m^{2} + n^{2}}{m + n}) - 2 m p$ $z = a = m^{2} + n^{2} - p^{2} - {(\frac{p (m - n) + m^{2} + n^{2}}{m + n})}^{2}$ $(x, y, z) i s p r i m i t i v e t r i p l e$ $S i m i l a r l y$ $z = b = 2 m q + 2 n p (z \in E)$ $. . . .$ $. . .$ $z = c = 2 n q - 2 m p (z \in E)$ $. . . . .$ $T h e s e f o r m u l a s o r n o t o f p r a c t i c a l$ $u s e . B e c a u s e {i t}^{'} s v e r y d i f f i c u l t t o$ $c h o o s e s u c h m, n & p t h a t q i s i n t e g e r$ $a n d (m, n, p, q) = 1 & o t h e r c o n d i t i o n s$ $f u l f i l l .$ $I m r a c t i c a l f a r m u l a s$
	Commented by Rasheed.Sindhi last updated on 01/Oct/19

	$I h o p e t h e s e f o r m u l a s c o v e r a l l$ $p o s s i b l e p r i m i t i v e s o l u t i o n s . N o n -$ $p r i m i t i v e s o l u t i o n s a r e i n t e g r a l$ $m u l t i p l e o f p r i m i t i v e s o l u t i o n s .$
	Commented by Prithwish sen last updated on 01/Oct/19

	$Sir the primitive quadruples which you$ $derived from the {lebesgue}^{'} s identity . Are$ $you sure that it covers all the possible$ $primitive quadruples ?$
	Commented by Prithwish sen last updated on 01/Oct/19

	$I think these are the nessecary but not the$ $sufficient condition .$
	Commented by Rasheed.Sindhi last updated on 01/Oct/19

	Commented by Rasheed.Sindhi last updated on 01/Oct/19

	$‘ ‘ T h u s, a l l p r i m i t i v e P y t h a g o r e a n q u a d r u p l e s u$ $a r e c h a r a c t e r i z e d b y {L e b s g u e}^{'} s i d e n t i t y . . .^{″}$ $S i r I m e a n t b y a b o v e t h a t a l l p o s s i b l e$ $p r i m i t i v e r o o t s c a n b e d e r i v e d . M a y b e$ $I a m w r o n g .$
	Commented by Prithwish sen last updated on 01/Oct/19

	$Actually sir I have some doubt to clear . By$ $the way you have done an outstanding work .$ $Excellent thinking .$
	Commented by Rasheed.Sindhi last updated on 03/Oct/19

	$THANKS i n a d v a n c e s i r! M y e m a i l$ $i s b e l o w :$ $r a s h e e d s i n d h i @ y a h o o . c o m$
	Commented by Prithwish sen last updated on 03/Oct/19

	$Sir I have already sentyou the pdf . If you$ $received that please give a feedback .$
	Commented by Prithwish sen last updated on 03/Oct/19

	$Sir actually I find something on net regarding$ $pythagorean quadruples very interesting . I will post it later .$ $Now I am little busy . So please wait .$
	Commented by Prithwish sen last updated on 03/Oct/19

	Commented by Prithwish sen last updated on 03/Oct/19

	Commented by Prithwish sen last updated on 03/Oct/19

	Commented by Rasheed.Sindhi last updated on 03/Oct/19

	$T h α n X P r i t h w i s h s e n s i r f o r t h i s$ $p r e c i o u s i n f o r m a t i o n! p l s h a r e w e b$ $u r l a l s o!$
	Commented by Prithwish sen last updated on 03/Oct/19

	$Sir I am not so good at using gadgets . If you$ $send your mail then I can send you the entire$ $pdf .$
	Commented by Rasheed.Sindhi last updated on 03/Oct/19

	$S i r I r e c i e v e d t h e p d f . I^{'} l l s t u d y i t .$ $T h a n k s a g a i n . W e s h o u l d t h i n k h o w$ $c a n w e r e l a t e p y t h a g o r e n q u a d r u p l e s$ $t o t h e q u e s t i o n a b o v e . M y a t t e m p t i n$ $t h i s c o n n e c t i o n n e a r l y f a i l e d . . .$ $P l s i r s t u d y m y f i r s t a n s w e r a g a i n .$ $I t h i n k f o r m u l a s t h e r e c a n b e u s e d$ $t o p r o d u c e a l l p o s s i b l e s o l u t i o n s o f$ $x y = (x + y) z$
	Commented by Prithwish sen last updated on 03/Oct/19

	$Excellent work sir . If all the possible$ $primitive solutions can be produced then$ $the all non primitive solutions also can be$ $produced . Great work Sir .$
	Commented by Rasheed.Sindhi last updated on 03/Oct/19

	$T h a n k s f o r E n c o u r a g i n g s i r!$
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