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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"

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    $\begingroup$ Hint: you'd have $(x-y)(x+y)=1002$. Think about parity. $\endgroup$ – lulu Dec 11 '17 at 21:04
  • $\begingroup$ @lulu I think $x$ and $y$ are rational, not necessarily integers. $\endgroup$ – alex.jordan Dec 11 '17 at 21:08
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    $\begingroup$ Do you mean rationals or integers $x,y$? $\displaystyle\left(\frac{503}2\right)^2-\left(\frac{499}2\right)^2=1002$. $\endgroup$ – Professor Vector Dec 11 '17 at 21:09
  • $\begingroup$ The book says rationals, weird. $\endgroup$ – Vinicius L. Deloi Dec 11 '17 at 21:11
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    $\begingroup$ @lulu Indeed - applying to $1\times n$ gives $\left(\frac{n+1}2\right)^2-\left(\frac{n-1}2\right)^2=n$ $\endgroup$ – Mark Bennet Dec 11 '17 at 21:18
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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 $x-y$.

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On rational number you can certainly find solutions.

On integer note that:

$$x^2 - y^2 = (x+y)(x-y)=1002=501\cdot 2$$ Since $x+y$ and $x-y$ have the same parity there not exists any solution (in the factorization 2 should compare with exponent 0 or even).

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