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Question Number 110299 by Lekhraj last updated on 28/Aug/20

	Answered by Her_Majesty last updated on 28/Aug/20

	$5 x + 5 y = 2 x y$ $\Rightarrow 5 ∣ x y \Rightarrow 5 ∣ x \lor 5 ∣ y$ $l e t x = 5 k$ $25 k + 5 y = 10 k y$ $y = \frac{5 k}{2 k - 1}$ $k = 1 \Rightarrow y = 5 \land x = 5 b u t x \neq y$ $k = 3 \Rightarrow y = 3 \land x = 15 \Rightarrow \sqrt{x + y} = 3 \sqrt{2}$ $o n l y s o l u t i o n b e c a u s e \frac{5 k}{2 k - 1} > 2 \forall k > 0$ $o f c o u r s e x = 3 \land y = 15 i s t h e 2^{n d} p o s s i b i l i t y$ $b u t i t a l s o l e a d s t o \sqrt{x + y} = 3 \sqrt{2}$
	Commented by Lekhraj last updated on 28/Aug/20

	$Thanks$
	Commented by udaythool last updated on 28/Aug/20
	$x+y=2xy/5⇒5k+y=2ky for some k∈Z\{0}⇒5+m=2km⇒m=±1,±5 m=1⇒k=3⇒x=15,y=3⇒(√(x+y))=±3(√2) m=−1⇒k=−2⇒x=−10,y=2⇒(√(x+y))=±2(√2)i m=5⇒k=1⇒x=5,y=5 But x≠y⇒m≠5 m=−5⇒k=0 which is not possible.$
	$x + y = 2 x y / 5 \Rightarrow 5 k + y = 2 k y for$ $some k \in Z ∖ {0} \Rightarrow 5 + m = 2 k m \Rightarrow m = \pm 1, \pm 5$ $m = 1 \Rightarrow k = 3 \Rightarrow x = 15, y = 3 \Rightarrow \sqrt{x + y} = \pm 3 \sqrt{2}$ $m = - 1 \Rightarrow k = - 2 \Rightarrow x = - 10, y = 2 \Rightarrow \sqrt{x + y} = \pm 2 \sqrt{2} i$ $m = 5 \Rightarrow k = 1 \Rightarrow x = 5, y = 5 But x \neq y \Rightarrow m \neq 5$ $m = - 5 \Rightarrow k = 0 which is not possible .$
	Commented by Her_Majesty last updated on 28/Aug/20

	$t h e q u e s t i o n s a y s ‘ ‘ d i s t i n c t p o s i t i v e {i n t e g e r s}^{″}$ $s o a s I s t a t e d, {t h e r e}^{'} s o n l y o n e s o l u t i o n$
	Commented by Rasheed.Sindhi last updated on 28/Aug/20

	$\sqrt{x + y} = \pm 3 \sqrt{2} i s n o t c o r r e c t, I t h i n k .$ $B e c a u s e t h e s y m b o l ‘ ‘ \sqrt{}^{″} i s$ $u s e d f o r p o s i t i v e s q u a r e r o o t$ $o n l y . F o r b o t h r o o t s ‘ ‘ \pm {\sqrt{}}^{″} i s$ $u s e d . (A n d f o r n e g a t i v e r o o t$ $‘ ‘ - {\sqrt{}}^{″} i s u s e d .)$
	Commented by Her_Majesty last updated on 28/Aug/20

	$\sqrt{x} i s a u n i q u e n u m b e r \forall x \in C$ $i f w e a s k f o r t h e s o l u t i o n o f x^{2} = y w e$ $w a n t a l l p o s s i b l e v a l u e s f o r x a n d g e t$ $x_{1} = - \sqrt{y} a n d x_{2} = \sqrt{y} . w e w r i t e x = \pm \sqrt{y}$ $b u t t h a t o b v i o u s l y c a n n o t m e a n$ $x = - \sqrt{y} \land x = + \sqrt{y} . w e w o u l d h a v e t o w r i t e$ $x = - \sqrt{y} x o r x = + \sqrt{y} (d i f f e r e n t s y m b o l s f o r$ $x o r i n u s e) b u t w e a r e t o o l a z y a n d w r i t e$ $x = - \sqrt{x} \lor x = + \sqrt{x} w h i c h i s w r o n g . . .$ $i t w o u l d m a k e a b s o l u t e l y n o s e n s e t o t h i n k$ $\sqrt{y} = \pm \sqrt{y} b e c a u s e t h e n x_{1} = \pm (- \sqrt{y}) a n d$ $x_{2} = \pm (+ \sqrt{y}) \Rightarrow x_{1} = \mp \sqrt{y} a n d x_{2} = \pm \sqrt{y} . . .$
	Commented by udaythool last updated on 28/Aug/20

	$Oh!$ $Yes, I {haven}^{'} t noticed . . .$
	Answered by Rasheed.Sindhi last updated on 28/Aug/20

	$\frac{1}{x} = \frac{2}{5} - \frac{1}{y} = \frac{2 y - 5}{5 y}$ $x = \frac{5 y}{2 y - 5}$ $A s x, y \in Z^{+}, y ⩾ 3$ $F o r y = 3, x = 15$ $A l s o f o r s y m m e t r y$ $x = 3, y = 15$ $\sqrt{x + y} = \sqrt{3 + 15} = 3 \sqrt{2}$ $C o n t i n u e$
	Commented by Lekhraj last updated on 28/Aug/20

	$Thanks$
	Answered by 1549442205PVT last updated on 29/Aug/20
	$(1/x)+(1/y)=(2/5)⇔((x+y)/(xy))=(2/5)⇔2xy=5(x+y) ⇔4xy−10(x+y)+25=25 ⇔(2x−5)(2y−5)=25 (1)⇒(2x−5 )is an odd divisor of 25 ⇒2x−5∈{1,−1,5,−5,−25,25} i)2x−5=1⇒x=3 .Replace into (1) we get y=15 ii)2x−5=−1⇒x=2.Replace into (1)we get y=−10 ∉Z^+ ⇒rejected iii)2x−5=5⇒x=5.Replace into (1) we get y=5 iv)2x−5=−25⇒x=−10⇒rejected v)2x−5=25⇒x=15⇒y=3 Combinating all above cases we get (x,y)∈{(3,15),(5,5),(15,3)} ⇒(√(x+y)) ∈{3(√2) ,(√(10))}$
	$\frac{1}{x} + \frac{1}{y} = \frac{2}{5} \Leftrightarrow \frac{x + y}{xy} = \frac{2}{5} \Leftrightarrow 2 xy = 5 (x + y)$ $\Leftrightarrow 4 xy - 10 (x + y) + 25 = 25$ $\Leftrightarrow (2 x - 5) (2 y - 5) = 25 (1) \Rightarrow (2 x - 5) is$ $an odd divisor of 25$ $\Rightarrow 2 x - 5 \in {1, - 1, 5, - 5, - 25, 25}$ $i) 2 x - 5 = 1 \Rightarrow x = 3 . Replace into (1) we$ $get y = 15$ $ii) 2 x - 5 = - 1 \Rightarrow x = 2 . Replace into$ $(1) we get y = - 10 \notin Z^{+} \Rightarrow rejected$ $iii) 2 x - 5 = 5 \Rightarrow x = 5 . Replace into (1)$ $we get y = 5$ $iv) 2 x - 5 = - 25 \Rightarrow x = - 10 \Rightarrow rejected$ $v) 2 x - 5 = 25 \Rightarrow x = 15 \Rightarrow y = 3$ $Combinating all above cases we get$ $(x, y) \in {(3, 15), (5, 5), (15, 3)}$ $\Rightarrow \sqrt{x + y} \in {3 \sqrt{2}, \sqrt{10}}$
	Commented by Her_Majesty last updated on 29/Aug/20

	$s a y i n g i t a g a i n, {w e}^{'} r e s e a r c h i n g f o r d i s t i n c t$ $i n t e g e r s \Rightarrow x \neq y$
	Commented by Rasheed.Sindhi last updated on 29/Aug/20

	$N i c e S i r!$ $O n l y e x c l u d e (5, 5) b e c a u s e t h e y$ $a r e n o t d i s t i n c t .$ ${Y o u}^{'} v e g i v e n t h e p r o b l e m u s e f u l l$ $f o r m :$ $(2 x - 5) (2 y - 5) = 25$ $T h i s f o r m o f f e r s l i m i t e d t r i a l s!$
	Commented by 1549442205PVT last updated on 30/Aug/20

	$Thank Sir . I {don}^{'} t note that$ $hypothesis .$
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