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[parent] Fermat's theorem proof (Proof)

Consider the sequence $ a,\,2a,\,\ldots,\,(p-1)a$.

They are all different (modulo $ p$), because if $ ma=na$ with $ 1\le m<n\le p-1$ then $ 0=a(m-n)$, and since $ p\nmid a$, we get $ p\mid(m-n)$, which is impossible.

Now, since all these numbers are different, the set $ \{a,\,2a,\,3a,\,\ldots,\,(p-1)a\}$ will have the $ p-1$ possible congruence classes (although not necessarily in the same order) and therefore

$\displaystyle a\cdot2a\cdot3a\cdots (p-1)a\equiv (p-1)!a^{p-1}\equiv (p-1)!\pmod{p}$
and using $ \gcd((p-1)!,\;p)=1$ we get
$\displaystyle a^{p-1}\equiv 1\pmod{p}.$



"Fermat's theorem proof" is owned by drini. [ full author list (2) | owner history (1) ]
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See Also: Euler-Fermat theorem, Fermat's little theorem, proof of Euler-Fermat theorem using Lagrange's theorem, inductive proof of Fermat's little theorem proof


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Cross-references: order, classes, congruence, numbers, sequence

This is version 6 of Fermat's theorem proof, born on 2001-10-15, modified 2008-01-30.
Object id is 223, canonical name is FermatsTheoremProof.
Accessed 8265 times total.

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AMS MSC: 11-00 (Number theory :: General reference works )
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