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.
Show answer
- D. \(a\) must not be divisible by \(p\)
Exercise · ZW-FERMAT-S-LITTLE-THEOREM-003 · ★☆☆ Basic
Choose one option.
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 \(p\).
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.
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