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  "body": "\n## Proof\n\nLet $\\Z[i]$ be the ring of [[Gaussian integers|0K5W69AC0J]] and let $N(a + b i) = a^2 + b^2$ be its [[norm|03AN1V4ZFJ]].\n\n1. For $z = a + b i$, let $\\bar z = a - b i$ be its complex conjugate. Then $N(z) = z \\bar z$.\n2. Complex conjugation is multiplicative: $\\overline{z w} = \\bar z \\, \\bar w$. So $N(z w) = z w \\bar z \\bar w = N(z) N(w)$ for all $z, w \\in \\Z[i]$.\n3. The elements $1, -1, i, -i$ lie in $\\Z[i]$. They are [[units|019NFGFPR5]], because $1 \\cdot 1 = 1$, $(-1)(-1) = 1$ and $i \\cdot (-i) = -i^2 = 1$.\n4. Conversely, let $u = a + b i$ be a unit of $\\Z[i]$, with $a, b \\in \\Z$. Then there is $v \\in \\Z[i]$ with $u v = 1$.\n5. By step 2, $N(u) N(v) = N(1) = 1$. Both $N(u)$ and $N(v)$ are nonnegative integers, so $N(u) = 1$.\n6. So $a^2 + b^2 = 1$ with $a, b \\in \\Z$. If $\\abs{a} \\ge 2$ or $\\abs{b} \\ge 2$, the sum is at least $4$. So $a, b \\in \\set{-1, 0, 1}$.\n7. The sum $a^2 + b^2$ is $1$ exactly when one of $a, b$ is $0$ and the other is $\\pm 1$. So $u \\in \\set{1, -1, i, -i}$.\n8. By steps 3 and 7, the units of $\\Z[i]$ are exactly $1$, $-1$, $i$ and $-i$.\n\n## Idea\n\nThe norm is multiplicative and takes nonnegative integer values. If $u v = 1$, the norms multiply to $1$, so $N(u) = 1$. The only lattice points at distance $1$ from the origin are $\\pm 1$ and $\\pm i$.\n",
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