Question-110299

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 = 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 \dots

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} \dots

Commented by udaythool last updated on 28/Aug/20

Oh!

Yes, I {haven}^{'} t noticed \dots

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

\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 .