ot\equiv 1 \), \( x

February 22, 2026

$ot\equiv 1 \), \( x$

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Solution: We seek the smallest integer \( x
ot\equiv \pm1 \pmod{17} \) such that \( x^4 \equiv 1 \pmod{17} \). The multiplicative order of \( x \) modulo 17 divides 16 (since \( \phi(17) = 16 \)), and since \( x^4 \equiv 1 \), the order divides 4.
The possible orders dividing 4 are 1, 2, or 4. We want \( x^4 \equiv 1 \), but \( x^2
ot\equiv \pm1 \), 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 \( x^4 \equiv 1 \pmod{17} \).

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