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Home » Solution: Let $x^2 \equiv 25 \pmod100$. - Abbey Badges

Solution: Let $x^2 \equiv 25 \pmod100$. - Abbey Badges

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 constant: \(b + B = 5 \Rightarrow 1 + B = 5 \Rightarrow B = 4\)

📌 So remainder is \(-2x + 4\)? But earlier substitution gave \(3\omega + 4\), contradiction.

📌 Wait â recalculate at \(\omega\):

📌 $\omega^2 = \omega - 1$

📌 $\omega^3 = \omega \cdot \omega^2 = \omega(\omega - 1) = \omega^2 - \omega = (\omega - 1) - \omega = -1$

📌 $x^4 = -\omega$

📌 $3x^3 = 3(-1) = -3$

📌 $-2x^2 = -2(\omega - 1) = -2\omega + 2$

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