Prove that there does not exists RATIONAL numbers $x$ and $y$ such that $x^2 - y^2 = 1002$ I tried to prove this using contradiction.
Suppose there does exists rationals $x$ and $y$ such that $x^2 - y^2 = 1002$
Let $ x = a/b  $ and $y = c/d$ 
Irreducible fractions.
I come up with $$ (ad)^2 - (bc)^2 = 1002 \cdot (bd)^2 $$
Thus
$$ (ad)^2 = (cb)^2 \pmod{1002}$$
And i'm stuck on this
Any tip or hint?
Thanks
This problem appears in the first chapter 'Contemporary Abstract Algebra"

If $(x^2 - y^2) = 1002$
$(x + y)(x - y) = 1002$
Let $x + y = 2$ and $x - y = 501$
Solve for $x,y$.   $x =\frac {503}2$ and $y = - \frac {499}{2}$.
That's a solution.  $\frac {503^2}4 - \frac {499^2}{4} = \frac {253009-249001}4=\frac {4008}{4 }= 1002$.
So not true.  
Now if $x$ and $y$ must be integers..
We have $(x+y)(x-y) = 2*3*167$.
$(x + y) + (x-y) = 2x$ so $x+y$ and $x-y$ must both be odd or both be even.  But $2$ must divide one but only one of $x+y$ or $x-y$.
