Comment of Gauss, on the even integer $x$, in the $p=x^2+y^2$, for primes of the form $p \stackrel{4}{\equiv} 1$

Question

$\color{Red}{\text{Question}}$: Let $p \stackrel{4}{\equiv} 1$ be a prime number. As you know, there exist natural numbers $x$ and $y$ such that $\color{Blue}{p=x^2+y^2}$; $\color{Red}{\text{ where $x$ is even }}$, and $y$ is odd.

Many years ago, I saw a result by Gauss, $\color{Red}{\text{ about the even number $x$ }}$ in $\color{Blue}{\text{ the above relation }}$, saying that $x$ satisfies a congruence condition:

$$x \stackrel{f(p)}{\equiv} g(p),$$

where $f(p)$ and $g(p)$ are functions only depending on $p$.

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Edit: Regarding the answer by user "@WhatsUp", it is better for the reader to consider this $\color{Green}{\text{ new question }}$ instead of the $\color{Red}{\text{ previous question}}$:

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$\color{Green}{\text{New Question}}$: Let $p \stackrel{4}{\equiv} 1$ be a prime number. As you know, there exist integers $x$ and $y$ such that $\color{Blue}{p=x^2+y^2}$; $\color{Green}{\text{ where $x\stackrel{4}{\equiv} 1$ }}$, and $y$ is even. It is obvious that $x$ is unique. (Also $y$ is unique up to $\pm$ sign.)

Many years ago, I saw a result by Gauss, $\color{Green}{\text{ about the number $x\stackrel{4}{\equiv} 1$ }}$ in $\color{Blue}{\text{ the above relation }}$, saying that $x$ satisfies a congruence condition:

$$x \stackrel{f(p)}{\equiv} g(p),$$

where $f(p)$ and $g(p)$ are functions only depending on $p$.

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I can not remember what are $f$ and $g$; but I guess that $f(p)=p$, and $g(p)$ was some function containing some factorials. Do anyone has ever seen this statemate?

Answer 1

I guess it's the following:

Let $p \equiv 1 \mod 4$ be a prime number, and let $a, b$ be integers such that $a^2 + b^2 = p$ and $a \equiv 1 \mod 4$. Then we have: $$a \equiv \frac{1}{2}\binom{\frac{p - 1}{2}}{\frac{p - 1}{4}}\mod p.$$

This appears in the determination of quartic Gauss sums, as is discussed in this paper. Note however that the minus sign is an obvious error in that paper.

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The above is however about the odd number. But the even number is not far: we may assume $b \equiv -\frac{p - 1}{2}!a \mod p$, as in that paper.

More directly, we have $b \equiv -\frac{1}{2}\left(\frac{\frac{p - 1}{2}!}{\frac{p - 1}{4}!}\right)^2 \mod p$. Note however that the $b$ here is not always positive, and hence could be either $x$ or $-x$ in the original question.

Thus we may only get the congruence $x \equiv \pm\frac{1}{2}\left(\frac{\frac{p - 1}{2}!}{\frac{p - 1}{4}!}\right)^2 \mod p$.