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Prove that a prime $p\equiv 1\pmod 8$ can be written in the form $x^2+2y^2.$

I referred to Show every prime $p\equiv 1,3$ (mod 8) can be written as $p=x^2+2y^2$

But it didn't give the solution I require.

We are supposed to use

break up the numbers $\{1, 2, . . . , p − 1\}$ into sets of the form $\{x, \bar{x}, -x, -\bar{x}, ix, −ix, i\bar{x}, −i\bar{x}\}$, where $i^2 \equiv −1 \pmod p$. Then analyze how many elements these sets have.

Here $\bar{x}$ mean the inverse of $x\pmod p$

So we have $\{1, 2, . . . , p − 1\}.$ I didn't get how to incorporate $i.$

So here's the analysis for $\{x, \bar{x}, -x, -\bar{x}\}.$

  • If $x\equiv \bar{x}\pmod p\implies x^2\equiv 1\pmod p\implies x\equiv 1,p-1.$

This gives ${1,p-1}$

  • If $x\equiv -{x}\pmod p\implies 2x\equiv 0\pmod p.$ Not possible.
  • If $x\equiv -\bar{x}\pmod p\implies x^2\equiv -1\pmod p.$

Which is possible only when $p\equiv 1\pmod 4.$ And this will give only $x,-\bar{x}$ to be different. That is $-x\equiv \bar{x}\pmod p.$

Now these break elements of $\{1, 2, . . . , p − 1\}$ into groups of $4$ or $2$ ( of ${1,p-1}$ and sometimes ${x,-\bar{x}}$)

Now, I did the same thing with $\{ix, −ix, i\bar{x}, −i\bar{x}\}.$

  • If $ix\equiv i\bar{x}\pmod p\implies x^2\equiv 1\pmod p\implies x\equiv 1,p-1.$
  • If $ix\equiv -{x}i\pmod p\implies 2x\equiv 0\pmod p.$ Not possible.
  • If $ix\equiv -\bar{x}i\pmod p\implies x^2\equiv -1\pmod p.$ This will give us only $x,-\bar{x}$ to be different.

Any hints?

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  • $\begingroup$ math.stackexchange.com/questions/1303441/… $\endgroup$
    – Asinomás
    Commented Aug 12, 2021 at 15:55
  • $\begingroup$ @Yorch it doesn't contain the answer i need.. :( $\endgroup$ Commented Aug 12, 2021 at 16:41
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    $\begingroup$ it appears you are asking about the first step only, that $-2$ is a quadratic residue mod an odd prime $p$ when $p \equiv 1 \pmod 8;$ still true when $p \equiv 3 \pmod 8$ of course. Some books I have do Gauss lemma and quadratic character of $2$ together. $\endgroup$
    – Will Jagy
    Commented Aug 12, 2021 at 16:44
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    $\begingroup$ Oh yes! Ic how qrs are coming in. Well yes, we can divide by y^2 and get (x/y)^2+2\equiv 0\mod p, which is true only for p, 1,3 mod 8. And we are done. But I want to use the hint, since this method the book had is supposed to use very elementary things. $\endgroup$ Commented Aug 12, 2021 at 16:52

1 Answer 1

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Your analysis is almost there. Take $p \equiv 1\pmod{8}.$

Look at $A =\{x, x^{-1}, -x, -x^{-1}\}.$ The only possible overlap is if $x^2 \equiv \pm 1\pmod{p},$ as you analyzed.

Now, look at $B=\{ix, ix^{-1}, -ix, -ix^{-1}\}.$ The only possible overlap is again if $x^2\equiv \pm 1\pmod{p}.$

When is $A \cap B$ non-empty? Well, if $x \in A\cap B$ then we must have $x \equiv \pm ix^{-1}\pmod{p}$ or $x^2 \equiv \pm i\pmod{p}.$ Similarly, if $x^{-1}, -x, -x^{-1}$ belong to $A\cap B$ you derive the same conclusion about $x.$

So, except for when $x$ is either $1, -1, i, -i,$ or a square root of $\pm i,$ the set $A\cup B$ has eight distinct elements. For $x=1,$ the set has elements $\{1, -1, i, -i\},$ which is four elements. With the zero leftover, this only gives us a number of elements which is 5 mod 8; in particular, since $p$ is 1 mod 8, we're missing some elements. Thus, there has to be some $x$ so that $x^2 \equiv \pm i\pmod{p}.$ Since $-1$ is a square mod $p,$ $i$ has a square root iff $-i$ does, so wlog say $x^2 \equiv i \pmod{8}.$

Now, in the complex numbers, it's easy to see that a square root of $i$ should look like $\sqrt{2}/2 + \sqrt{-2}/2.$ This is the motivation for what we do.

Because now that we know $x^2 \equiv i\pmod{8},$ to extract a square root of $2$ we just consider $2x/(1+i).$ Squaring this we get $4x^2/(1+i)^2 = 4x^2/2i = -2ix^2 = -2ii = 2.$ Thus, $2$ has a square root modulo $p.$

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