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In a post Efficiently finding two squares which sum to a prime

I read

"In 1825 Gauss gave the following construction for writing a prime congruent to $1 \pmod{4}$ as a sum of two squares: Let $p=4k+1$ be a prime number. Determine $x$ (this is uniquely possible...) so that

$$ x = \frac{(2k)!}{2(k!)^2} \pmod{p}, \quad |x| < \frac{p}{2}$$

Now determine $y$ so that

$$ y = x \cdot (2k)! \pmod{p}, \quad |y| < \frac{p}{2}$$

Gauss showed that $x^2+y^2=p$."

I checked the Stark's book and I did not see a direct reference to the Gauss original paper or book or other reference to explain the method how to come to this and find the formula. If you know the reference or you can explain its procedure, I will be grateful.

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asad
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We have $$ (2k)!^2 \equiv (-1)^{2k}(4k)! = (p-1)! \equiv -1 \pmod p $$ by writing $-1 \equiv 4k \pmod p$, $-2\equiv 4k-1 \pmod p$, etc. so that $(2k)! \equiv (-1)^{2k} (4k)(4k-1)\cdots(2k+1) \pmod p$. Then $$ x^2 \equiv \frac{(2k)!^2}{4(k!)^4} \equiv \frac{-1}{4(k!)^4} \pmod p. $$ Similarly, $y^2 \equiv \frac{1}{4(k!)^4} \pmod p$. So $x^2 + y^2 \equiv 0 \pmod p$. Now I am guessing that Gauss brilliantly bounded $x^2+y^2$ using his choice of $x$ and $y$ but the trivial bound is $$ x^2 + y^2 = |x|^2 + |y|^2 < \frac{p^2}4 + \frac{p^2}4 = \frac{p^2}2. $$ So I am guessing Gauss did something smart ; I checked the book referenced in your link and it doesn't redirect to a proof, the author only says that Gauss proved that.

Hope that helps,

Patrick Da Silva
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    "So I am guessing Gauss did something smart" – Well, Gauss was smart :-) – lhf Nov 10 '15 at 10:02
  • @Patrick Da Silva, Thanks for verification, but my main question was how he (or one can) construct this neat formula. – asad Nov 10 '15 at 13:50
  • @asad : Yeah sorry, I got that part of the question,;the point of my answer is to tell you it's a hard question. Maybe some experienced number theorist remembers Gauss' proof from reading it somewhere. – Patrick Da Silva Nov 10 '15 at 15:29