Difference between revisions of "Fermat's Little Theorem"
IntrepidMath (talk | contribs) |
(→Sample Problem) |
||
Line 11: | Line 11: | ||
=== Sample Problem === | === Sample Problem === | ||
− | 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 | + | 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> |
By Fermat's Little Theorem, we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5. Hence,<br> | By Fermat's Little Theorem, we know <math>{n^{5}}</math> is congruent to <math>n</math> [[modulo]] 5. Hence,<br> |
Revision as of 14:37, 18 June 2006
Contents
[hide]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.
Sample Problem
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.