# Fermat test

Gold Member
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 wich 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.

Hurkyl
Staff Emeritus
Gold Member
The statement

ap - a is a multiple of p

is equivalent to

ap = 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.

Gold Member
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?

Hurkyl
Staff Emeritus
Gold Member
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.

Gold Member
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?

Gold Member
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?

Hurkyl
Staff Emeritus
Gold Member
Because

ap = a (mod p)

iff

(a+kp)p = a+kp (mod p)

P.S. all those little p's are supposed to be exponents.

Gold Member
still dont get it )-: how does it prooves a+kp<p?

sorry to bother you.

Hurkyl
Staff Emeritus