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📅 May 4, 2026 👤 scraface

May 04, 2026

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\( 11^4 = (121)^2 \equiv (121 - 7\cdot17 = 121 - 119 = 2)^2 = 4
\( 12^4 = (144)^2 \equiv (144 - 8\cdot17 = 144 - 136 = 8)^2 = 64 \equi...
\( 13^4 = (169)^2 \equiv (16)^2 = 256 \equiv 256 - 15\cdot17 = 256 - 2...
\( 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...
Wait: \( 4^4 = 256 \equiv 1 \), \( 13^4 = 28561 \), but earlier comput...
But \( 4, 13 \), and since group is cyclic, primitive roots â actual...
But we need order 4. \( 4^2 = 16 \equiv -1

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