fermat test

MathematicalPhysicist

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

The statement

a^p - a is a multiple of p

is equivalent to

a^p = 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.

MathematicalPhysicist

why is it that?

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.

MathematicalPhysicist

x is integer?

MathematicalPhysicist

excuse my ignorance but how does this proove you only need to check values of a smaller than p?

Hurkyl

Because

a^p = a (mod p)

iff

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


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

MathematicalPhysicist

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


sorry to bother you.

Hurkyl

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).