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📅 March 11, 2026 👤 scraface

Mar 11, 2026

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Subtracting equation (1) from (2):
[x^2 + (y - 1)^2 + z^2] - [(x - 1)^2 + y^2 + z^2] = 0
x^2 + y^2 - 2y + 1 + z^2 - (x^2 - 2x + 1 + y^2 + z^2) = 0
-2y + 2x = 0 \quad \Rightarrow \quad x = y
Similarly, subtract (1) from (3):
[x^2 + y^2 + (z - 1)^2] - [(x - 1)^2 + y^2 + z^2] = 0
x^2 + y^2 + z^2 - 2z + 1 - (x^2 - 2x + 1 + y^2 + z^2) = 0
-2z + 2x = 0 \quad \Rightarrow \quad x = z
So $x = y = z$. Plug into equation (1):
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