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Let $p$ be a prime with $p\equiv 3\pmod 8$ , then all rational solutions to $px^4-4y^4=z^2$ satisfy $xyz=0$.

Apparently, the argument to prove this is that $-1$ is not a quadratic residue of $p$.

I know that $-1$ is not a quadratic residue of $p$ because of the law of quadratic reciprocity and $p\equiv 3mod4$. However, I fail to see how this proves that all solutions satisfy xyz=0. If I take the equation modulo $p$, then I'm stuck with $-4y^4\equiv z^2 mod p$. What am I missing?

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  • $\begingroup$ If $z$ is nonzero modulo $p$, then $(2y^2/z)^2\equiv -1\pmod p$. $\endgroup$
    – richrow
    Nov 2, 2021 at 7:16

1 Answer 1

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Assuming that $x,y,z$ are non-zero, it's clearly equivalent to the following : $px^4=4y^2+z^2=(2y^2)^2+z^2$. Hence $px^4$ can be written as sum of two perfect squares, which is a contradiction noting that $p\overset{4}{\equiv}3$ and $\nu_p(px^4)\overset{2}{\equiv}1$. For more information on the theorem check the following link:
https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem

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  • $\begingroup$ What is $V_p(px^4)$? $\endgroup$
    – Leli
    Nov 2, 2021 at 5:13
  • $\begingroup$ A term in p-adic calculations. Greatest positive integer $k$ such that $p^k|px^4$ but $p^{k+1}\not|px^4$ $\endgroup$
    – Aryan
    Nov 2, 2021 at 5:15
  • $\begingroup$ Okay, thank you. This helps a lot ! $\endgroup$
    – Leli
    Nov 2, 2021 at 5:16
  • $\begingroup$ @Aryan FYI, the $p$-adic order function is usually written as $\nu$, i.e., \nu, a Greek letter, which you can also confirm by checking the Wikipedia article's text. Also, it seems from your comment that you're using a vertical bar for stating an expression divides another expression. This affects spacing & how \not is handled. You got "$p^k | px^4$ but $p^{k+1} \not | px^4$". However, replacing | with \mid gives "$p^k \mid px^4$ but $p^{k+1} \not\mid px^4$". $\endgroup$ Nov 2, 2021 at 5:20
  • $\begingroup$ @JohnOmielan Thanks a lot for your writing notes, I didn't know about the second one, and I also didn't know that LaTeX has the Greek symbol $\endgroup$
    – Aryan
    Nov 2, 2021 at 5:23

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