number-theory-Solve-in-Z-1-x-1-y-1-xy-1-4-

Question Number 142514 by mnjuly1970 last updated on 01/Jun/21

\dots . . n u m b e r t h e o r y \dots . .

S o l v e i n Z :

\frac{1}{x} + \frac{1}{y} + \frac{1}{x y} = \frac{1}{4} \dots . ?

\dots \dots \dots

Answered by ArielVyny last updated on 01/Jun/21

c o n d i t i o n s x \neq 0 y \neq 0 x y \neq 0

\frac{1}{x} = t \frac{1}{y} = u

t + u + t u = \frac{1}{4}

t (1 + u) = \frac{1}{4} - u

t = - \frac{u - \frac{1}{4}}{1 + u}

- \frac{u - \frac{1}{4}}{1 + u} + u - (\frac{u - \frac{1}{4}}{1 + u}) \times u = \frac{1}{4}

- \frac{4 u - 1}{1 + u} + u - (\frac{4 u - 4}{1 + u}) \times u = 1

- \frac{4 u - 1}{1 + u} + \frac{u (1 + u)}{1 + u} - (\frac{4 u^{2} - 4 u}{1 + u}) = 1

- 4 u + 1 + u (1 + u) - (4 u^{2} - 4 u) = 1 + u

- 4 u + 1 + u + u^{2} - 4 u^{2} + 4 u = 1 + u

- 3 u^{2} = 0 u = 0

t = - \frac{u - \frac{1}{4}}{1 + u}

t = - \frac{- \frac{1}{4}}{1} t = \frac{1}{4} = \frac{1}{x}

Answered by ArielVyny last updated on 01/Jun/21

i t i s p o s s i b l e ?

Answered by ArielVyny last updated on 01/Jun/21

i t h i n k i s n o t p o s s i b l e

Answered by ajfour last updated on 01/Jun/21

\frac{1}{x} + \frac{1}{y} + \frac{1}{x y} = \frac{1}{4}

\Rightarrow 4 x + 4 y + 4 = x y

(x - 4) (y - 4) = 20

= 1 \times 20 = 2 \times 10 = 4 \times 5

\Rightarrow (x, y) = (5, 24), (6, 14), (8, 9)

Commented by mnjuly1970 last updated on 01/Jun/21

t h s n k s a l o t \dots

(2, - 6), (6, - 2), (- 16, 3), (3, - 16)

f o r e x a m p l e :

(2, - 6)

\frac{1}{2} - \frac{1}{6} - \frac{1}{12} = \frac{6 - 2 - 1}{12} = \frac{1}{4} . . ✓

Answered by mnjuly1970 last updated on 01/Jun/21

s o l u t i o n :

\frac{y + x + 1}{x y} = \frac{1}{4}

4 x + 4 y + 4 - x y = 0

x (4 - y) = - 4 y - 4

x = \frac{- 4 y - 4}{4 - y} = \frac{4 (y + 1)}{y - 4} = 4 . \frac{y - 4 + 5}{y - 4}

= 4 + \frac{20}{y - 4} \dots . x, y \in Z

y - 4 = \pm 1, y - 4 = \pm 2, y - 4 = 4

y - 4 = 5, y - 4 = \pm 10

y - 4 = \pm 20

, \dots \dots e a s y \dots