I've started a little reading on quadratic reciprocity, and a reason for this has eluded me. Here's a little of what I came up with so far. I decided I want to show that for all primes $p$, if $p|x^2-2$, then $p$ does not divide $2y^2+3$. Then, by way of contradiction, if $(x^2-2)/(2y^2+3)$ is an integer, then any $p$ such that $p|2y^2+3$ would have to divide $x^2-2$, a contradiction. I see this is true for $p=2$. I want to find all $p$ such that $x^2\equiv 2\pmod{p}$, and since for any odd $p$,


I see $(2|p)=1$ iff $p\equiv 1,7\pmod{8}$. So only primes of the form $8k+1$ or $8k+7$ divide $x^2-2$. However, I don't see a way to show that primes of the form $8k+1$ or $8k+7$ do not divide $2y^2+3$, so maybe I'm completely off the mark. Does anyone know how to resolve this, or have a better idea of what to do? Thanks!

  • $\begingroup$ no wonder you can't see it; it's false. Take y = 2. (I'm reading the problem as 2y^2 + 3, although you wrote 2y^3 + 3 a few times; I assume this was a typo.) $\endgroup$ – Qiaochu Yuan Oct 24 '10 at 9:25
  • $\begingroup$ Whoops, you're right, I'll fix that right away. Thanks also for your response, I'll attempt that now. $\endgroup$ – yunone Oct 24 '10 at 9:37
  • 2
    $\begingroup$ +1: For an interesting question and showing your prior work. $\endgroup$ – Aryabhata Oct 24 '10 at 13:16

Try to show that $2y^2 + 3$ must have at least one prime divisor which is not of the form $8k+1$ or $8k+7$.

  • 2
    $\begingroup$ Thanks, I realized that $2y^2+3$ is always of the form $8k+3$ or $8k+5$, but that products of terms of the form $8k+1$ or $8k+7$ stay in the same form. $\endgroup$ – yunone Oct 24 '10 at 22:33

Let's suppose it is an integer;

$x^2 - 2 = k(2y^2+3)$

$x^2 = (2k)y^2 + 3k + 2;$

since $x$ is an integer; l.h.s is also a perfect square, which is a quadratic equation in $y$;

that means roots of the equation are equal, and discriminant = 0;

$b^2-4ac = 0$ $\ \Rightarrow\ $ $0-(3k + 2)(2K) = 0\ $; $\ \Rightarrow\ $ $k = -3/2$ or $k = 0$;

and $k = 0$ only for $x^2 = 2$;

which is a contradiction...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.