- Hi Everybody:

Fermat says that if A is a PRIME integer, then 2^A divided by A will

leave a remainder of 2.

It occurred to me that perhaps we might not have to raise 2 all the way

to exponent A, but only to some fractional part of A and still have just

as good a probable prime test as Fermat while significantly reducing

computation.

I came up with 2 statements listed below. They seem to be true but I

can't prove them. I chased Little Fermat all over the web looking for

something like this but with no luck. I'm starting to suspect that maybe

I'm all wet, that I've run into something that works only for small

numbers. Can anyone help?

Here are the 2 statements, either of which (if true) could be used as the

basis of a probable prime test:

If A is a PRIME integer, then 2^((A+1)/2) divided by A will leave a

remainder of 2 OR a remainder of (A-2).

If A is a PRIME integer, then 2^((A-1)/2) divided by A will leave a

remainder of 1 OR a remainder of (A-1).

I would appreciate any comments. My thanks and regards to all.

Bill Sindelar - Bill Sindelar wrote:

> If A is a PRIME integer, then 2^((A-1)/2) divided by A will leave a

You are absolutely correct about this, although you must add the

> remainder of 1 OR a remainder of (A-1).

condition that A be an ODD integer.

This is the initial step toward coming up with the concept used for

defining "strong probable primes" -- there is plenty written about

this and easily accessible with a web search, but you didn't know

what to look for (strong probable prime).

Start with:

http://www.utm.edu/research/primes/prove/prove2_3.html - On Sat, 05 May 2001, William F Sindelar wrote:
> Hi Everybody:

Bill,

> Fermat says that if A is a PRIME integer, then 2^A divided by A will

> leave a remainder of 2.

> It occurred to me that perhaps we might not have to raise 2 all the way

> to exponent A, but only to some fractional part of A and still have just

> as good a probable prime test as Fermat while significantly reducing

> computation.

> I came up with 2 statements listed below. They seem to be true but I

> can't prove them. I chased Little Fermat all over the web looking for

> something like this but with no luck. I'm starting to suspect that maybe

> I'm all wet, that I've run into something that works only for small

> numbers. Can anyone help?

> Here are the 2 statements, either of which (if true) could be used as the

> basis of a probable prime test:

> If A is a PRIME integer, then 2^((A+1)/2) divided by A will leave a

> remainder of 2 OR a remainder of (A-2).

> If A is a PRIME integer, then 2^((A-1)/2) divided by A will leave a

> remainder of 1 OR a remainder of (A-1).

> I would appreciate any comments. My thanks and regards to all.

> Bill Sindelar

You might want to look into (modular) exponentiation algorithms, as the calculations

x^n (mod p)

x^((n-1)/2) (mod p)

take almost the same number of steps, in fact only one square and one multiply are required to get from the latter to the former.

A rule of thumb is that an m-bit exponent, n, takes roughly

m squares and m/2 multiplies on average. Here m =~ lg(n) the binary logarithm (round it up).

Phil

Mathematics should not have to involve martyrdom;

Support Eric Weisstein, see http://mathworld.wolfram.com

Find the best deals on the web at AltaVista Shopping!

http://www.shopping.altavista.com - Hi there Bill

>Here are the 2 statements, either of which (if true) could be used as

the

>basis of a probable prime test:

What you've discovered is the basis for two slight improvements on

>If A is a PRIME integer, then 2^((A+1)/2) divided by A will leave a

>remainder of 2 OR a remainder of (A-2).

>If A is a PRIME integer, then 2^((A-1)/2) divided by A will leave a

>remainder of 1 OR a remainder of (A-1).

Fermat probable primality. Euler's test does precisely as you describe,

and looks at b^((A-1)/2). As you've discovered, it's either +1 or -1

mod A, if A is prime, since it must be a square root of +1. (In fact,

you can determine which result should occur before embarking on the

calculation, and this is sufficient to prove Proth primes, given the

right choice of b).

This is a little less naive than Fermat PRP, and help catch a few

'obvious' pseudoprimes. A little more work will give the Miller-Rabin

test - see http://www.utm.edu/research/primes/prove/prove2_3.html. It's

also the beginnings of the N-1 primality testing methods, where you

consider factors of N-1 other than 2. http://www.utm.edu/research/

primes/prove/prove3_1.html is a gentle introduction to this - notice

the proofs all revolve around "the order of a, modulo p" - the smallest

integer e>0 such that a^e=1 mod p.

As far as execution times of these tests go, the calculation of a^b mod

c can be done in time that scales with the number of bits in b - so

saving one bit in b does not save much. On the other hand, the amount

of extra calculation involved in the theorems on Chris C's page is not

great, so this line of thought can see very strong probable primality

results with not too much effort - and ultimately, deterministic proofs

for those forms where we can factor N-1.

Chris N.