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. - Veritas Home Health