But \( 4, 13 \), and since group is cyclic, primitive roots — 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.

Mon, 04 May 2026 02:51:43 +0000

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