Vieta jumping is awesome!

Number Theory Level 2

Let x x and y y be positive integers such that x y ∣ x 2 + y 2 + 1. xy|x^2+y^2+1. Find the value of x 2 + y 2 + 1 x y \dfrac{x^2+y^2+1}{xy}


The answer is 3.

View wiki
Rohit Udaiwal by Rohit Udaiwal , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Anweshan Bor
Nov 28, 2015

put x=y=1.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

What if x=1, y=2 or x=2, y=5

Sachin Sharma - 5 years, 6 months ago

Log in to reply

We are given, x y xy divides x 2 + y 2 + 1 x^2 + y^2 +1 . Now the obtained questient is 1 x + 1 y + 1 x y \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{xy} . Note that this is a whole number only when the individual terms are. And the only possible case is x = 1 ; y = 1 x=1; y=1

Manish Mayank - 5 years, 5 months ago

Log in to reply

I think it should be

x 2 + y 2 + 1 x y = x y + y x + 1 x y \dfrac{x^2+y^2+1}{xy}\ = \dfrac{x}{y}\ + \dfrac{y}{x}\ + \dfrac{1}{xy}

and now check what happens if ( x , y ) = ( 1 , 2 ) (x,y)=(1,2) or ( 2 , 5 ) (2,5)

Sachin Sharma - 5 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...