# Fermat's Little Theorem

## Theorem

Let $p$ be a prime number.

Let $n \in \Z_{>0}$ be a positive integer such that $p$ is not a divisor of $n$.

Then:

- $n^{p - 1} \equiv 1 \pmod p$

### Corollary 1

If $p$ is a prime number, then $n^p \equiv n \pmod p$.

### Corollary 2

Let $p$ be a prime number.

Then:

- $n^{p - 1} \equiv \sqbrk {p \nmid n} \pmod p$

where:

- $\nmid$ denotes non-divisibility
- $\sqbrk \cdots$ is Iverson's convention.

### Corollary 3

Let $p^k$ be a prime power for some prime number $p$ and $k \in \Z_{\gt 0}$.

Then:

- $\forall n \in \Z_{\gt 0}: n^{p^k} \equiv n \pmod p$

### Corollary 4

Let $p^k$ be a prime power for some prime number $p$ and $k \in \Z_{\gt 0}$.

Let $n \in \Z_{\gt 0}$ with $p \nmid n$.

Then:

- $n^{p^k - 1} \equiv 1 \pmod p$

## Proof 1

Consider the integer sequence $n, 2 n, 3 n, \dotsc, \paren {p - 1} n$.

Note that none of these integers is congruent modulo $p$ to any of the others.

If this were the case, we would have $a n \equiv b n \pmod p$ for some $1 \le a < b \le p - 1$.

Then as $\map \gcd {n, p} = 1$, and we can cancel the $n$, we get $a \equiv b \pmod p$ and so $a = b$.

We have that:

- $\forall c \in \set {1, 2, \dotsc, p - 1}: p \nmid n \land p \nmid c$

So by Euclid's Lemma:

- $p \nmid c n$

for any such $c n$, which means:

- $c n \not \equiv 0 \pmod p$

Thus, each integer in the sequence can be reduced modulo $p$ to exactly one of $1, 2, 3, \ldots, p - 1$.

So $\set {1, 2, 3, \ldots, p - 1}$ is the set of Reduced Residue System modulo $p$.

So, upon taking the product of these congruences, we see that:

- $n \times 2 n \times 3 n \times \dots \times \paren {p - 1} n \equiv 1 \times 2 \times 3 \times \cdots \times \paren {p - 1} \pmod p$

This simplifies to:

- $n^{p - 1} \times \paren {p - 1}! \equiv \paren {p - 1}! \pmod p$

Since $p \nmid \paren {p - 1}!$, we can cancel $\paren {p - 1}!$ from both sides, leaving us with:

- $n^{p - 1} \equiv 1 \pmod p$

$\blacksquare$

## Proof 2

By Prime not Divisor implies Coprime:

- $p \nmid n \implies p \perp n$

and Euler's Theorem can be applied.

Thus:

- $n^{\map \phi p} \equiv 1 \pmod p$

But from Euler Phi Function of Prime Power:

- $\map \phi p = p \paren {1 - \dfrac 1 p} = p - 1$

and the result follows.

$\blacksquare$

## Proof 3

Let $\struct {\Z'_p, \times}$ denote the multiplicative group of reduced residues modulo $p$.

From the corollary to Reduced Residue System under Multiplication forms Abelian Group, $\struct {\Z'_p, \times}$ forms a group of order $p - 1$ under modulo multiplication.

By Element to Power of Group Order is Identity, we have:

- $n^{p - 1} \equiv 1 \pmod p$

$\blacksquare$

## Proof 4

Proof by induction over $n$.

Induction base:

- $1^p \equiv 1 \pmod p$

Induction step:

Assume $n^p \equiv n \pmod p$

\(\ds \paren {n + 1}^p\) | \(=\) | \(\ds \sum_{k \mathop = 0}^p {p \choose k} n^{p - k} \cdot 1^k\) | Binomial Theorem | |||||||||||

\(\ds \forall k: 0 < k < p: \, \) | \(\ds {p \choose k}\) | \(\equiv\) | \(\ds 0 \pmod p\) | Binomial Coefficient of Prime |

and so:

\(\ds \sum_{k \mathop = 0}^p {p \choose k} n^{p - k}\) | \(\equiv\) | \(\ds n^p + n^0 \pmod p\) | ||||||||||||

\(\ds \) | \(\equiv\) | \(\ds n^p + 1 \pmod p\) | ||||||||||||

\(\ds \) | \(\equiv\) | \(\ds n + 1 \pmod p\) | Induction Step |

Dividing by $n$:

- $\forall n: n^p \equiv n \pmod p \implies n^{p - 1} \equiv 1 \pmod p$

$\blacksquare$

## Examples

### $12$ Divides $n^2 - 1$ if $\gcd \set {n, 6} = 1$

- $12$ divides $n^2 - 1$ if $\gcd \set {n, 6} = 1 \implies 12 \divides n^2 - 1$

## Also known as

Some sources call this **Fermat's Theorem**, but it needs to be appreciated that this may cause confusion with Fermat's Last Theorem.

## Also defined as

Some sources refer to the corollary $n^p \equiv n \pmod p$ as **Fermat's little theorem** and from it derive this result.

## Source of Name

This entry was named for Pierre de Fermat.

## Historical Note

Fermat's Little Theorem was first stated, without proof, by Pierre de Fermat in $1640$.

Chinese mathematicians were aware of the result for $n = 2$ some $2500$ years ago.

The appearance of the first published proof of this result is the subject of differing opinions.

- Some sources have it that the first published proof was by Leonhard Paul Euler $1736$.

- Others state that it was first proved by Gottfried Wilhelm von Leibniz in an undated manuscript, and that he may have known a proof before $1688$, perhaps as early as $1683$.

- MathWorld's page on the subject reports the first published proof as being by Leonhard Paul Euler $1749$, but it is possible this has been conflated with the proof for Fermat's Two Squares Theorem.

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.6$ - 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\text {3-2}$ Fermat's Little Theorem: Exercise $1$ - 1978: John S. Rose:
*A Course on Group Theory*... (previous) ... (next): $0$: Some Conventions and some Basic Facts: Exercise $8$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 19$: Properties of $\Z_m$ as an algebraic system (mention) - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 2.3$: Congruences: Theorem $3$ - 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $64$ - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.13$: Fermat ($\text {1601}$ – $\text {1665}$) - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $64$ - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next): Entry:**Fermat's theorem** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**Fermat's theorem** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**Fermat's Little Theorem**