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Home » rac{\sqrt3}4s^2 = 25\sqrt3 - Midis

rac{\sqrt3}4s^2 = 25\sqrt3 - Midis

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 \( 13^4 = (169)^2 \equiv (16)^2 = 256 \equiv 256 - 15\cdot17 = 256 - 255 = 1 \) â yes

📌 \( 14^4 = (-3)^4 = 81 \equiv 13

📌 \( 15^4 = (-2)^4 = 16

📌 \( 16^4 = (-1)^4 = 1 \), but \( 16 \equiv -1 \), excluded

📌 So solutions to \( x^4 \equiv 1 \pmod{17} \) are \( x \equiv 4, 13, 13, \ldots \) â wait: \( 4, 13 \), and also check \( 16 \equiv -1 \): \( (-1)^4 = 1 \), but excluded.

📌 Wait: \( 4^4 = 256 \equiv 1 \), \( 13^4 = 28561 \), but earlier computations show both â¡ 1.

📌 But \( 4, 13 \), and since group is cyclic, primitive roots â actually, the solutions are \( x \equiv g^{0}, g^{4}, g^{8}, g^{12} \) where \( g \) is a primitive root. But easier: since \( 4^4 = 256 \equiv 1 \), \( 13^2 = 169 \equiv 16 \equiv -1 \), so \( 13^4 = (-1)^2 = 1 \). So yes, \( 4, 13 \) are solutions.

📌 But we need order 4. \( 4^2 = 16 \equiv -1

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