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  "body": "\n## Proof\n\nWe expand both sides, using the laws of a [[commutative ring|0GYGHKX834]].\n\n1. The left side is $(a^2 + b^2)(c^2 + d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2$.\n2. We have $(ac - bd)^2 = a^2c^2 - 2abcd + b^2d^2$ and $(ad + bc)^2 = a^2d^2 + 2abcd + b^2c^2$.\n3. Adding the two lines of step 2, the terms $\\pm 2abcd$ cancel. The sum is $a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2$, which is the left side by step 1.\n4. Next, $(ac + bd)^2 = a^2c^2 + 2abcd + b^2d^2$ and $(ad - bc)^2 = a^2d^2 - 2abcd + b^2c^2$.\n5. Adding the two lines of step 4, the terms $\\pm 2abcd$ cancel again. The sum is again the left side.\n6. Steps 3 and 5 prove both equalities.\n\n## Idea\n\nEach square on the right gives two of the four products $a^2c^2, a^2d^2, b^2c^2, b^2d^2$ and a cross term $\\pm 2abcd$. The signs are chosen so that the cross terms cancel. Over the complex numbers this is $\\abs{z}^2 \\abs{w}^2 = \\abs{zw}^2$ for $z = a + bi$ and $w = c + di$.\n",
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