Exercise · ZW-FERMAT-S-LITTLE-THEOREM-003 · ★☆☆ Basic

If \(p\) is prime, what condition on \(a\) is needed to use \(a^{p-1}\equiv 1 \pmod{p}\)?

Choose one option.

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  • D. \(a\) must not be divisible by \(p\)

Explanation

The necessary hypothesis is \(p\nmid a\). Requiring \(a\) to be even or prime is not a condition of the theorem, and neither is requiring \(a\) to be greater than \(p\).

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Review the concept: Fermat's little theorem

Fermat's little theorem states that if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then \(a^{p-1}\equiv 1 \pmod{p}\). It is a basic tool for simplifying modular powers.

\[a^{p-1}\equiv 1 \pmod{p}, \quad p\nmid a\]

More practice problems: fermat's little theorem

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