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