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Second proof. This proof is based on the properties of the quartic residue symbol in $\mathbb{Z}[i]$. I will rely on Chapter 9 of Ireland-Rosen: A classical introduction to modern number theory (2nd ed.), especially Theorem 2, Proposition 9.8.5, and Proposition 9.8.6 there. Assume that $p\equiv 1\pmod{16}$. There are 8 pairs representing $p$, hence we can assume that our pair $(A,B)$ is the one with $$A\equiv 1\pmod{8}\qquad\text{and}\qquad B\equiv 0\pmod{4}.$$ As $A/B$ is a square-root of $-1$ modulo $p$, it has order four modulo $p$, hence it is actually a fourth power modulo $p$. Therefore, it suffices to show that $A$ is a fourth power modulo $p$. Equivalently, $A$ is a fourth power in $\mathbb{Z}[i]/(A+Bi)$. However, this follows from the law of quartic reciprocity and the conditions on $(A,B)$: $$\left(\frac{A}{A+Bi}\right)_4=\left(\frac{A+Bi}{A}\right)_4 =\left(\frac{Bi}{A}\right)_4=\left(\frac{i}{A}\right)_4=(-1)^{(A-1)/4}=1.$$

Second proof. This proof is based on the properties of the quartic residue symbol in $\mathbb{Z}[i]$. I will rely on Chapter 9 of Ireland-Rosen: A classical introduction to modern number theory (2nd ed.), especially Theorem 2, Proposition 9.8.5, and Proposition 9.8.6 there. Assume that $p\equiv 1\pmod{16}$. There are $8$ pairs representing $p$, hence we can assume that our pair $(A,B)$ is the one with $$A\equiv 1\pmod{8}\qquad\text{and}\qquad B\equiv 0\pmod{4}.$$ As $A/B$ is a square-root of $-1$ modulo $p$, it has order four modulo $p$, hence it is actually a fourth power modulo $p$. Therefore, it suffices to show that $A$ is a fourth power modulo $p$. Equivalently, $A$ is a fourth power in $\mathbb{Z}[i]/(A+Bi)$. However, this follows from the law of quartic reciprocity and the conditions on $(A,B)$: $$\left(\frac{A}{A+Bi}\right)_4=\left(\frac{A+Bi}{A}\right)_4 =\left(\frac{Bi}{A}\right)_4=\left(\frac{i}{A}\right)_4=(-1)^{(A-1)/4}=1.$$

The first observation follows from the law of quadratic reciprocity. Indeed, assume that $p\equiv 1\pmod{8}$ and $p=A^2+B^2$. Let $A'$ denote the odd part of $A$. Then $p\equiv B^2\pmod{A'}$, and hence $\left(\frac{p}{A'}\right)=1$. It follows that $\left(\frac{A'}{p}\right)=1$, and then by $\left(\frac{2}{p}\right)=1$ also $\left(\frac{A}{p}\right)=1$. In the same way, $\left(\frac{B}{p}\right)=1$. The second observation lies deeper. I will give two proofs, the first one giving more information than we need.

Second proof. I will use Gerry Myerson's helpful comment below the original post. For any prime $p=4n+1$, Gauss (1825) observed that $p=A^2+B^2$, where $A$ and $B$ are the absolute least residues modulo $p$ of $(2n)!/(2n!^2)$ and $(2n)!^2/(2n!^2)$. For a proof, see Jacobsthal (1907) or the "Added" section below. Hence it suffices to show that in case when $4$ divides $n$, these two integers are fourth powers modulo $p$. So let us assume that $4\mid n$. By Wilson's theorem, $$(2n)!^{(p-1)/4}=(2n)!^{n}\equiv(p-1)!^{n/2}\equiv(-1)^{n/2}=1\pmod{p},$$ hence $(2n)!$ is a fourth power modulo $p$. So we only need to prove that $2n!^2$ is a fourth power modulo $p$. Equivalently, $$2^{(p-1)/4}n!^{(p-1)/2}\equiv 1\pmod{p}.\tag{1}$$ However, this follows from Theorem 5 in Emma Lehmer's 1977 paper "Generalizations of Gauss's lemma", and we are done.

The first observation follows from the law of quadratic reciprocity. Indeed, assume that $p\equiv 1\pmod{8}$ and $p=A^2+B^2$. Let $A'$ denote the odd part of $A$. Then $p\equiv B^2\pmod{A'}$, and hence $\left(\frac{p}{A'}\right)=1$. It follows that $\left(\frac{A'}{p}\right)=1$, and then by $\left(\frac{2}{p}\right)=1$ also $\left(\frac{A}{p}\right)=1$. In the same way, $\left(\frac{B}{p}\right)=1$. The second observation lies deeper. I will give three proofs (which actually yield more than we need).

Second proof. This proof is based on the properties of the quartic residue symbol in $\mathbb{Z}[i]$. I will rely on Chapter 9 of Ireland-Rosen: A classical introduction to modern number theory (2nd ed.), especially Theorem 2, Proposition 9.8.5, and Proposition 9.8.6 there. Assume that $p\equiv 1\pmod{16}$. There are 8 pairs representing $p$, hence we can assume that our pair $(A,B)$ is the one with $$A\equiv 1\pmod{8}\qquad\text{and}\qquad B\equiv 0\pmod{4}.$$ As $A/B$ is a square-root of $-1$ modulo $p$, it has order four modulo $p$, hence it is actually a fourth power modulo $p$. Therefore, it suffices to show that $A$ is a fourth power modulo $p$. Equivalently, $A$ is a fourth power in $\mathbb{Z}[i]/(A+Bi)$. However, this follows from the law of quartic reciprocity and the conditions on $(A,B)$: $$\left(\frac{A}{A+Bi}\right)_4=\left(\frac{A+Bi}{A}\right)_4 =\left(\frac{Bi}{A}\right)_4=\left(\frac{i}{A}\right)_4=(-1)^{(A-1)/4}=1.$$

Third proof. I will use Gerry Myerson's helpful comment below the original post. For any prime $p=4n+1$, Gauss (1825) observed that $p=A^2+B^2$, where $A$ and $B$ are the absolute least residues modulo $p$ of $(2n)!/(2n!^2)$ and $(2n)!^2/(2n!^2)$. For a proof, see Jacobsthal (1907) or the "Added" section below. Hence it suffices to show that in case when $4$ divides $n$, these two integers are fourth powers modulo $p$. So let us assume that $4\mid n$. By Wilson's theorem, $$(2n)!^{(p-1)/4}=(2n)!^{n}\equiv(p-1)!^{n/2}\equiv(-1)^{n/2}=1\pmod{p},$$ hence $(2n)!$ is a fourth power modulo $p$. So we only need to prove that $2n!^2$ is a fourth power modulo $p$. Equivalently, $$2^{(p-1)/4}n!^{(p-1)/2}\equiv 1\pmod{p}.\tag{1}$$ However, this follows from Theorem 5 in Emma Lehmer's 1977 paper "Generalizations of Gauss's lemma", and we are done.

First proof. I will use some results of Gauss, Jacobi, and Eisenstein collected in Section III.13.5 of Hasse's 1965 book: Bericht über neuere Untersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper, Teil II: Reziprozitätsgesetz. Assume that $p\equiv 1\pmod{16}$. Without loss of generality, $A$ is odd and $B$ is even. As $A/B$ is a square-root of $-1$ modulo $p$, it has order four modulo $p$, hence it is also a fourth power modulo $p$. Therefore, it suffices to show that $B$ is a fourth power modulo $p$. Let us write $B=2^k B'$, where $B'$ is odd. As $k\geq 2$, every prime divisor of $B'$ is a fourth power modulo $p$. Hence $B'$ is a fourth power modulo $p$, and it remains to show that $2^k$ is a fourth power modulo $p$. If $k=2$, then we are done as $2$ is a quadratic residue modulo $p$. If $k\geq 3$, then we are done as $2$ is itself a fourth power modulo $p$.

First proof. I will use some results of Gauss, Jacobi, and Eisenstein collected in Section III.13.5 of Hasse's 1965 book: Bericht über neuere Untersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper, Teil II: Reziprozitätsgesetz. Assume that $p\equiv 1\pmod{16}$. Without loss of generality, $A$ is odd and $B$ is even. As $A/B$ is a square-root of $-1$ modulo $p$, it has order four modulo $p$, hence it is actually a fourth power modulo $p$. Therefore, it suffices to show that $B$ is a fourth power modulo $p$. Let us write $B=2^k B'$, where $k\geq 2$ and $B'$ is odd. By the results in Hasse's book, every prime divisor of $B'$ is a fourth power modulo $p$, hence $B'$ is a fourth power modulo $p$. It remains to show that $2^k$ is a fourth power modulo $p$. If $k\geq 3$, then $2$ is a fourth power modulo $p$ (see Hasse's book), hence we are done. If $k=2$, then we are also done upon noting that $2$ is a quadratic residue modulo $p$.