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The possible orders dividing 4 are 1, 2, or 4. We want \( x^4 \equiv 1 \), but \( x^2
ot\equiv 1 \), \( x
ot\equiv \pm1 \), so order exactly 4.
The multiplicative group modulo 17 is cyclic of order 16. The number of solutions to \( x^4 \equiv 1 \pmod{17} \) is \( \gcd(4,16) = 4 \)? No â actually, the number of solutions to \( x^d \equiv 1 \pmod{p} \) is \( \gcd(d, p-1) \), so here \( \gcd(4,16) = 4 \). So there are 4 solutions.
We find all \( x \in \{2,3,\dots,16\} \) such that \( x^4 \equiv 1 \pmod{17} \).
Try values:
\( 2^4 = 16 \equiv -1
ot\equiv 1 \)
\( 3^4 = 81 \equiv 81 - 4\cdot17 = 81 - 68 = 13
\( 4^4 = (16)^2 = (-1)^2 = 1 \mod 17 \) â yes