Difference between revisions of "Fermat's Little Theorem"
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A frequently used corolary of Fermat's little theorem is <math> a^p \equiv a \pmod {p}</math>. | A frequently used corolary 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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| + | === Example === | ||
| + | One of Euler's conjectures was disproved in then 1960s by three American mathematicians when they showed there was a positive integer such that <math>133^5+110^5+84^5+27^5=n^5</math>. Find the value of <math>{n}</math>. ([[AIME]] 1989 #9)<br><br> | ||
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| + | By Fermat's Little Theorem, we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5. Hence,<br> | ||
| + | <center><math>3 + 0 + 4 + 7 \equiv n\pmod{5}</math></center><br> | ||
| + | <center><math>4 \equiv n\pmod{5}</math></center><br> | ||
| + | Continuing, we examine the equation modulo 3, | ||
| + | <center><math>-1 + 1 + 0 + 0 \equiv n\pmod{3}</math></center><br> | ||
| + | <center><math>0 \equiv n\pmod{3}</math></center><br> | ||
| + | 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 13:14, 18 June 2006
Contents
Statement
If 
is an integer and 
is a prime number, then 
.
Note: This theorem is a special case of Euler's totient theorem.
Corollary
A frequently used corolary of Fermat's little theorem is 
.
As you can see, it is derived by multipling both sides of the theorem by a.
Example
One of Euler's conjectures was disproved in then 1960s by three American mathematicians when they showed there was a positive integer such that 
. Find the value of 
. (AIME 1989 #9)
By Fermat's Little Theorem, we know 
is congruent to 
modulo 5. Hence,
Continuing, we examine the equation modulo 3,
Thus, is divisible by three and leaves a remainder of four when divided by 5. It's obvious that 
so the only possibilities are 
or 
. It quickly becomes apparent that 174 is much too large so must be 144.
Credit
This theorem is credited to Pierre de Fermat.
