Difference between revisions of "Fermat's Little Theorem"

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=== Corollary ===
 
=== Corollary ===
  
A frequently used corolary of Fermat's little theorem is <math> a^p \equiv a \pmod {p}</math>.
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A frequently used corollary of Fermat's little theorem is <math> a^p \equiv a \pmod {p}</math>.
 
As you can see, it is derived by multipling both sides of the theorem by a.
 
As you can see, it is derived by multipling both sides of the theorem by a.
  
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<br><br><br>
 
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Thus, <math>n</math> is divisible by three and leaves a remainder of four when divided by 5.  It's obvious that <math>n>133</math> so the only possibilities are <math>n = 144</math> or <math>n = 174</math>.  It quickly becomes apparent that 174 is much too large so <math>n</math> must be 144.
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Thus, <math>n</math> is divisible by three and leaves a remainder of four when divided by 5.  It's obvious that <math>n>133</math>, so the only possibilities are <math>n = 144</math> or <math>n = 174</math>.  It quickly becomes apparent that 174 is much too large, so <math>n</math> must be 144.
  
 
=== Credit ===
 
=== Credit ===

Revision as of 13:05, 26 June 2006

Fermat's Little Theorem is highly useful in number theory for simplifying computations in modular arithmetic (which students should study more at the introductory level if they have a hard time following the rest of this article).


Statement

If ${a}$ is an integer and ${p}$ is a prime number, then $a^{p-1}\equiv 1 \pmod {p}$.

Note: This theorem is a special case of Euler's Totient Theorem.

Corollary

A frequently used corollary of Fermat's little theorem is $a^p \equiv a \pmod {p}$. As you can see, it is derived by multipling both sides of the theorem by a.

Sample Problem

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that $133^5+110^5+84^5+27^5=n^5$. Find the value of ${n}$. (AIME 1989/9)

By Fermat's Little Theorem, we know ${n^{5}}$ is congruent to $n$ modulo 5. Hence,

$3 + 0 + 4 + 7 \equiv n\pmod{5}$
$4 \equiv n\pmod{5}$



Continuing, we examine the equation modulo 3,

$-1 + 1 + 0 + 0 \equiv n\pmod{3}$
$0 \equiv n\pmod{3}$




Thus, $n$ is divisible by three and leaves a remainder of four when divided by 5. It's obvious that $n>133$, so the only possibilities are $n = 144$ or $n = 174$. It quickly becomes apparent that 174 is much too large, so $n$ must be 144.

Credit

This theorem is credited to Pierre de Fermat.

See also

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