# All rational solutions to $px^4-4y^4=z^2$ with $p\equiv 3\pmod8$ satisfy $xyz=0$

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?

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

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:
• What is $V_p(px^4)$?
• A term in p-adic calculations. Greatest positive integer $k$ such that $p^k|px^4$ but $p^{k+1}\not|px^4$ Nov 2, 2021 at 5:15
• @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 $p^{k+1} \not\mid px^4$". Nov 2, 2021 at 5:20