Let-x-y-Z-if-x-2-y-2-divide-xy-1-then-prove-x-2-y-2-xy-1-is-square-of-integer-number-

Question Number 53618 by kaivan.ahmadi last updated on 23/Jan/19

Let x, y \in Z

if x^{2} + y^{2} divide xy + 1 then prove

\frac{x^{2} + y^{2}}{xy + 1} is square of integer number

Commented by mr W last updated on 24/Jan/19

i t h i n k t h e r e i s o n l y o n e p o s s i b i l i t y :

x = y = 1 o r - 1

Commented by kaivan.ahmadi last updated on 24/Jan/19

thank you, chek the proof \dots .

Commented by Kunal12588 last updated on 28/Mar/19

l e t x = 2 a n d y = 8

\frac{2^{2} + 8^{2}}{(2) (8) + 1} = \frac{4 + 64}{16 + 1} = \frac{68}{17} = 4 = 2^{2}

Answered by kaivan.ahmadi last updated on 24/Jan/19

suppose that x = n \neq 0 then \frac{y^{2} + n^{2}}{ny + 1}

the root of ny + 1 = 0 must be the root

of y^{2} + n = 0

ny + 1 = 0 \Rightarrow y = \frac{- 1}{n} \in Z \Rightarrow n = \pm 1 \Rightarrow y = \pm 1, x = \pm 1

if x = 1 \Rightarrow y = 1

if x = - 1 \Rightarrow y = - 1

so \frac{x^{2} + y^{2}}{xy + 1} = 1 .

Now suppose that x = 0 then

\frac{x^{2} + y^{2}}{xy + 1} = y^{2} and there is no thing to proof .

similary y = 0

is it true mr w ?

Commented by mr W last updated on 24/Jan/19

{i t}^{'} s c o r r e c t s i r . t h i s i s m y t r y :

\frac{x y + 1}{x^{2} + y^{2}} = n = s o m e i n t e g e r > 0

\Rightarrow x y + 1 = n (x^{2} + y^{2}) ⩾ n (2 x y)

\Rightarrow (2 n - 1) x y ⩽ 1

\Rightarrow \frac{1}{2 n - 1} ⩾ x y ⩾ 1

\Rightarrow 2 n - 1 = 1

\Rightarrow n = 1 a n d x y = 1

\Rightarrow x = y = 1 (o r - 1)

Commented by kaivan.ahmadi last updated on 24/Jan/19

very nice . attaching x = 0 or y = 0 complete the proof .

thank you sir .