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📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📝 Recent Posts

x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1),
so $ x^6 \equiv 1 \pmod{d(x)} $. Thus, modulo $ d(x) $, all powers of...
We reduce $ f(x) \equiv (x^2 + 1)^{2025} + 1 \pmod{d(x)} $.
Letâs analyze $ (x^2 + 1)^{2025} \mod d(x) $.
Let $ y = x^2 $, then $ d(x) = y^2 + y + 1 $. Since $ y^3 \equiv 1 \mo...
Now $ x^2 \equiv y $, so $ x^2 + 1 \equiv y + 1 $, and
We reduce $ (y + 1)^{2025} $ modulo $ y^2 + y + 1 $. Since $ y^2 \equi...
Letâs compute the reduction of $ (y + 1)^{2025} $ modulo $ y^2 + y +...
e 1 $, and $ \omega^2 + \omega + 1 = 0 $), and note that this polynomi...
We evaluate $ (y + 1)^{2025} $ at $ y = \omega $

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