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Thus, the number of successful additional attempts is - GetMeFoodie

📅 May 2, 2026 👤 scraface

May 02, 2026

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📌 \( 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

📌 e 1 \), so order 4. \( 13^2 = -1 \), \( 13^4 = 1 \), so both have order 4.

📌 Now check: \( 4

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