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Given p(x,y) = x2 - 7y2 - 24 ∈ ℤ[x,y], does the image in ℤ7[x,y] become x2 - 3 or x2 + 4? Or could I use either to determine whether or not p(x,y) has any solutions in ℤ[x,y]?

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    $\begingroup$ Either will do, they are equivalent $\pmod 7$. $\endgroup$
    – lulu
    Dec 10 '18 at 18:51
  • $\begingroup$ Indeed, @lulu, I would have said that as polynomials over the field with seven elements, they are equal. $\endgroup$
    – Lubin
    Dec 11 '18 at 4:11
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You could use either $x^2-3$ or $x^2+4$, or in fact $x^2+(4+n)\times 7$ for any $ n \in \mathbb N;$

adding any multiple of $7$ doesn't change things modulo $7$.

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