11Infinte descent settles this easily....
We have (ad)2 +(bc)2=3(bd)2 for a,b,c,d integers...
From the above eqn we have 3|(ad)2and 3|(bc)2....meaning 9 divides them...
So let (ad)2=9k2and (bc)2=9l2
From there we have 3|(bd)2....same argument 9|(bd)2
So we have (bd)2=9p2
So new eqn becomes k2+l2=3p2....Continuing this process indefinitely proves no existance of rationals....
I must thank Prophet sir, who helped me understand I.D...
