Integral solutions of a 2 + b 2 = a 2 b 2 a^2 + b^2 = a^2 b^2

Find the number of pairs of nonnegative integers ( a , b ) (a,b) satisfying

a 2 + b 2 = a 2 b 2 . a^2 + b^2 = a^2 b^2.

0 2 1 4

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

Ossama Ismail
Jan 24, 2017

Answer is 1

a 2 + b 2 = a 2 b 2 a^2 + b^2 = a^2 b^2

a 2 b 2 a 2 b 2 = 0 a^2 b^2 -a^2 - b^2 =0

a 2 b 2 a 2 b 2 + 1 = 1 a^2 b^2 -a^2 - b^2+1 =1

a 2 ( b 2 1 ) ( b 2 1 ) = 1 a^2 (b^2 - 1) - (b^2 -1) = 1

( a 2 1 ) ( b 2 1 ) = 1 (a^2-1) (b^2 - 1) = 1

The only integral solution to the above equation is a = b = 0 a =b=0

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...