Fermat's little theorem

  • Thread starter kmeado07
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  • #1
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Homework Statement



From fermat's little theorem deduce that when p is prime,

n^p is equivalent to n (mod p)

for all integers n.

Homework Equations





The Attempt at a Solution



I know from Fermat's Little Theorem that ,

n^(p-1) is equivalent to 1 (mod p),

but i don't know how to use it for this particular question.
 

Answers and Replies

  • #2
Dick
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If a is equivalent to b mod(p) then c*a is equivalent to c*b mod(p). What might be a good choice for c in your problem?
 
  • #3
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ok, so in my question, c=n ?
So by dividing by n i would get,

1^p is equiavlent to 1 (mod p)

How do i reach n^(p-1) on the left hand side?
 
  • #4
Dick
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n^p divided by n IS NOT 1^p. PLEASE stop and review algebra with exponents before you try to continue.
 
  • #5
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sorry, silly mistake.
dividing by n would give me,

n^(p-1) equivalent to 1 (mod p)

which is fermat's little theorem. so is this all i need to do?
 
  • #6
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This is ok if p doesnt divide n, but the question asks me for all integers n.
So how do i show it's also true for when p divides n?
 
  • #7
Dick
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This is ok if p doesnt divide n, but the question asks me for all integers n.
So how do i show it's also true for when p divides n?

You actually want to go the other way. Start with Fermat's little theorem and then go to your conclusion. Multiply by n, you can always do that.
 

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