Let $ y = x^2 $, then $ d(x) = y^2 + y + 1 $. Since $ y^3 \equiv 1 \mod (y^2 + y + 1) $, we have $ y^3 \equiv 1 $, so powers of $ y $ reduce modulo 3.

Tue, 05 May 2026 23:21:27 +0000

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