Fermat’s theorem proof

Consider the sequence $a,\mathrm{\hspace{0.17em}2}a,\mathrm{\dots },(p-1)a$.

They are all different (modulo $p$), because if $ma=na$ with 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  $\{a,\mathrm{\hspace{0.17em}2}a,\mathrm{\hspace{0.17em}3}a,\mathrm{\dots },(p-1)a\}$  will have the $p-1$ possible congruence classes (although not necessarily in the same order) and therefore

$$a\cdot 2a\cdot 3a\mathrm{\cdots }(p-1)a\equiv (p-1)!a^{p-1}\equiv (p-1)!(modp)$$

and using  $\gcd ((p-1)!,p)=1$  we get

$$a^{p-1}\equiv 1(modp).$$

Title	Fermat’s theorem proof
Canonical name	FermatsTheoremProof
Date of creation	2013-03-22 11:46:10
Last modified on	2013-03-22 11:46:10
Owner	drini (3)
Last modified by	drini (3)
Numerical id	11
Author	drini (3)
Entry type	Proof
Classification	msc 11-00
Classification	msc 37B55
Related topic	EulerFermatTheorem
Related topic	FermatsLittleTheorem
Related topic	ProofOfEulerFermatTheoremUsingLagrangesTheorem
Related topic	FermatsLittleTheoremProofInductive