Fermat's little theorem states that if [itex]p[/itex] is a prime number, then for any integer [itex]a[/itex], [itex]a^{p}-a[/itex] will be divisible by

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Equations

[tex]a^{p-1}\equiv1\pmod p \quad (\text{for\ }a \not\equiv 0 \pmod p)[/tex]

Fermat's Little Theorem
If p is a prime number and a an integer, then
[tex]a^p\equiv a\ (p)[/tex]

In order to prove Fermat's Little theorem, we will start by proving a superficially slightly weaker result, which is also referred to as Fermat's Little Theorem, on occasion. The two results imply each other, however.

Theorem
Let a and p be coprime, then

[tex]a^{p-1}-1 \equiv 0\ (p).[/tex]

Proof
Start by listing the first p-1 positive multiples of a:

[tex]a, 2a, 3a, \ldots, (p -1)a[/tex]

Suppose that [itex]ra[/itex] and [itex]sa[/itex] are the same modulo [itex]p[/itex], with [itex]0 <r,s < p[/itex]. Since [itex]a[/itex] is nonzero mod [itex]p[/itex], we can cancel, giving [itex]r \equiv s\ (p)[/itex]. So the [itex]p-1[/itex] multiples of [itex]a[/itex] above are distinct and nonzero; that is, they must be congruent to [itex]1, 2, 3, \ldots, p-1[/itex] in some order. Multiply all these congruences together and we find

Remark
This result can be proven by appeal to Lagrange's theorem, since the non-zero residues form a group modulo [itex]p[/itex]. Although we haven't proven, they are a group, we are explicitly using that multiplicative inverses modulo [itex]p[/itex] exist, which is an elementary application of Euclid's algorithm.

Corollary
Let [itex]p[/itex] be a prime and [itex]a[/itex] any integer, then [itex]a^p \equiv a\ (p)[/itex]. Proof
The result is trival (both sides are zero) if [itex]p[/itex] divides [itex]a[/itex]. If not then we need only multiply the congruence in Fermat's Little Theorem by [itex]a[/itex] to complete the proof.

Done the tex'ing. Removed (unneccessary and unused) reference to there being a minimal d with some property. Made it clearer what each 'proof' was proving.