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Home » To fix, lets change the initial condition to make a crossover possible. - Malaeb

To fix, lets change the initial condition to make a crossover possible. - Malaeb

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 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

📌 ot\equiv \pm1 \), \( 13 \equiv -4 \), so also not \( \pm1 \). So both valid? But are there only two?

📌 Wait, earlier list missed: \( (-4)^2 = -1 \), so \( (-4)^4 = 1 \), \( -4 \equiv 13 \), already have.

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Modulo 125, terms with 125 or higher vanish:
Check divisibility by higher numbers.
Among any 4 consecutive integers, there are:
One multiple of 4, and we already have factor of 8
Solution: Since $ g(x) $ is a cubic polynomial, we assume:
Using the given values, we form the system:

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