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We reduce $ (y + 1)^{2025} $ modulo $ y^2 + y + 1 $. Since $ y^2 \equiv -y - 1 $, we can reduce all powers of $ y $ to linear expressions.
Let’s compute the reduction of $ (y + 1)^{2025} $ modulo $ y^2 + y + 1 $. Let’s denote $ \omega $ as a primitive cube root of unity (so $ \omega^3 = 1 $, $ \omega
e 1 $, and $ \omega^2 + \omega + 1 = 0 $), and note that this polynomial divides $ y^2 + y + 1 $.

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