Fermat's Test for Prime Numbers and Pseudo Primes: Explained

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SUMMARY

The discussion centers on Fermat's Test for determining prime numbers, as described in Paul Hoffman's book "The Man Who Loved Only Numbers." The test states that if n is prime, then for every whole number a, the expression a^n - a is a multiple of n. However, pseudo prime numbers, such as 561, can deceive this test. The conversation highlights that due to modulo arithmetic, it is sufficient to check values of a that are less than p to determine if n is a pseudo prime.

PREREQUISITES
  • Understanding of Fermat's Little Theorem
  • Basic knowledge of modulo arithmetic
  • Familiarity with prime and pseudo prime numbers
  • Concept of mathematical induction
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  • Research advanced primality testing algorithms, such as the Miller-Rabin test
  • Explore the properties and applications of pseudo primes in number theory
  • Study the implications of Fermat's Little Theorem in cryptography
  • Learn about the relationship between Fermat's Test and other primality tests
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Mathematicians, computer scientists, and students of number theory interested in prime number testing and the properties of pseudo primes.

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i read the book "the man who loved only numbers" by paul hoffman, and there is explanation about fermat's test for checking prime numbers which states: if n is prime then for every whole number a the number a^n-a is a multiple of n.
now my question is about a pseduo prime number which "fools" this test how could you find such a number, i mean you should check every a which is ofcourse infinite numbers how could you possibly know that n is pseduo prime number by not checking every a?

btw the smallest p.p number is 561.
 
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The statement

ap[/size] - a is a multiple of p

is equivalent to

ap[/size] = a (modulo p)


So, by the virtue of modulo arithmetic, we only need to check values of a that are less than p.

There are probably much more efficient theoretical tests for pseudoprimes, but my number theory book is at work so I can't look them up.
 
Originally posted by Hurkyl



So, by the virtue of modulo arithmetic, we only need to check values of a that are less than p.

why is it that?
 
I'll presume it's obvious that px = 0 mod p.


Consider this:

(a+kp)b = ab + kpb = ab (mod p)

Letting b = (a+kp)i, you can prove by induction that

(a+kp)n = an for all n.
 
Originally posted by Hurkyl
I'll presume it's obvious that px = 0 mod p.


Consider this:

(a+kp)b = ab + kpb = ab (mod p)

Letting b = (a+kp)i, you can prove by induction that

(a+kp)n = an for all n.
x is integer?
 
Originally posted by Hurkyl


(a+kp)n = an for all n.
excuse my ignorance but how does this proove you only need to check values of a smaller than p?
 
Because

ap[/size] = a (mod p)

iff

(a+kp)p[/size] = a+kp (mod p)


P.S. all those little p's are supposed to be exponents.
 
still don't get it )-: how does it prooves a+kp<p?


sorry to bother you.
 
It doesn't.

Once we've checked all values of a less than p, this theorem extends our results to any other value of a because any integer n can be written as m + pk for some integer m in [0, p).
 

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