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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 $p$,

$$\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$$

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!

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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.) –  Qiaochu Yuan Oct 24 '10 at 9:25
    
Whoops, you're right, I'll fix that right away. Thanks also for your response, I'll attempt that now. –  yunone Oct 24 '10 at 9:37
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+1: For an interesting question and showing your prior work. –  Aryabhata Oct 24 '10 at 13:16
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up vote 7 down vote accepted

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

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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. –  yunone Oct 24 '10 at 22:33
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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...

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