FLT and Binomials
 

:unsure:
Is there a method of utilising Fermat's little theorem and the Binomial theorem and factorising large numbers. Is it well known and widely used?

Mally :coffee:

An algorithm for primes.
 

:smile:
An Algorithm for determining primes.
Since I have not received any response on this thread I give below a method to determine primes. Kindly examine it for whatever merit it contains and whether it can be successfully used for determining primes.
First of all a ready table of primes should be available at our disposal. Lets say we have one up to 2^30.
For simplicity lets determine the primality of a known prime say 641.

Using FLT then 2^(641-1)-1 is completely divisible by 641.
Hence 2^640 – 1 = (2^320 +1) ( 2^320 - 1 )
Now 2^320 = (2^32)^10
= [(1073741824) x 2^2]^10 Because 2^30 = 1073741824 (available).
Let us express 17373741824 in units of 641 or in multiples of 641 as we are testing the divisibility of the expression by 641.
Now 1073741824 = (1675104)x 641 + 160.
Let us denote ALL the multiples of 641 by the letter N. As we proceed N will take different values which for simplicity we will consider as N all the time.
Therefore 1073741824 = [(N + 160.) x 2^2 ]^10
=(4N + 640)^10. BY simplifying the expression we get (N+N-1)^10.
Since N +N is a multiple this boils down to (N - 1)
Therefore 2^320 + 1 = (N-1)^10 + 1
2^320 – 1 = (N – 1)^10 -1
Hence (2^640) – 1 = (N -1)^20 – 1.
Now all the terms of the binomial expansion except the last term has N as part of the coefficient. So all these terms are divisible by 641.
The last term is 1^20 which cancels with -1 out side the expansion.
So we conclude that 2^640 – 1 is divisible by 641.
Therefore 641 is a prime number.

Mally :coffee:

[QUOTE=mfgoode;92796]:smile:
An Algorithm for determining primes.

For simplicity lets determine the primality of a known prime say 641.
[/QUOTE]
When trying to prove the primality of N, wouldn't this method require the factorization of N-1?

Also, wouldn't the last line of your proof be fooled by Carmichael numbers?

jasonp

Factorisation of N-1
 

[QUOTE=jasonp;92799]When trying to prove the primality of N, wouldn't this method require the factorization of N-1?

Also, wouldn't the last line of your proof be fooled by Carmichael numbers?

jasonp[/QUOTE]

To your 1st question NO! thats the beauty of the algorithm.

To your second, Im not sure but I woud say probably, as its based on the FLT and if you can fool FLT so can you do so by Carmichael numbers.

Good questions ! jasonp.

Mally :coffee::

Another worked example
 

:rolleyes:
Another worked example.

Since my last example was a bit laborious and confusing I give another example.

Let us test whether 524287 is a prime number or not. If it is then it must satisfy
2^(524287 -1) = 1 mod 524287.

Now from prime tables 2^19 = 524288.
If we express this no. in terms of our number 524287 (call it N in this case )

We have 2^19 = N + 1 (Where N is 524287 or a multiple , here it is 1x 524287)

Now 2^ (524287 – 1) – 1 = 2 ^ (524286) -1

This is (2^ 262143 + 1) (2^262143 – 1) [ a^2 – b^2 = (a+b)(a-b) ]

The exponent 262143 = (2^19)^ 13797 = (N + 1) ^13797

Because when 262143 is divided by 19 the answer is 13797.

There fore (2^19)^13797 + 1 = (N + 1) ^ 13797 + 1

(2^19)^13797 – 1 = (N + 1) ^ 13797 - 1

Therefore (2^19)^13797 = (N + 1)^ 13797

By squaring the above expression and subtracting 1

Therefore 2^ 524286 - 1 = (N + 1) ^ 27594 – 1 [524286 = 19 x 13797 x 2 ]

= (N + 1) ^27594 - 1

The last term of the binomial expansion is + 1 (from N + 1) and this cancels with – 1 in the expression itself

All the other terms of (N +1) ^27594 have N as part of the coefficients and hence are

divisible by 524287
.
So we have 2 ^ (524287 – 1 ) = = 1 mod 524287. Hence 524287 is a prime as per the FLT.

Mally :coffee:

primality /ne factoring
 

[QUOTE=mfgoode;93082]:rolleyes:
Another worked example.

Since my last example was a bit laborious ...
Hence 524287 is a prime as per the FLT.

Mally :coffee:[/QUOTE]

Not to quibble, but you might have better luck if this thread
were moved to "math" or somewhere else more appropriate.
Primality (proofs) are a completely different topic (runs in
polyn time, most likely polyn of degree 4, definitely degree
4 for expected runtime); while factoring is believed to be
hard [presuming P /ne NP]. In any case, best known method
is subexponential, for factoring. That would be gnfs, cf.
RSA155=RSA512, and the more recent RSA200.

Likewise, if we were considering primality here, the current
record for proof of a (random) prime is 20,000 digits (up from
15,000, both by the same person/group as RSA200). You don't
seem to be addressing questions of runtime, of the type needed
to scale your proposed method up to the current range of interest.
-bd (dept math, lehigh)

Re-direction of thread
 

[QUOTE=bdodson;93093]Not to quibble, but you might have better luck if this thread
were moved to "math" or somewhere else more appropriate.
QUOTE]
:smile:
Thank you bdodson for the suggestion of redirecting my thread to the math forum. Typical beaucracy! Moderator please do the needful.

I am not sure what you mean by 'better luck'

I am just an amateur mathematician and by no means an all rounder. As such I have no knowledge of programming and have no desire of ever attaining it.

Have you gone thru my worked examples to realise the potentiality of it?

I have merely given an algortihm and if you dont consider it as such lets call it a mathematical curiousity.

It is not the creation of a crank as I have resurrected this method by a respected Indian PhD who may have gone into obscurity had I not written about it. ( name witheld)

The reason why I have submitted it is to explore the possibility of working out a much shorter version as compared to the existing systems.

I would add that if you can represent M44 then there is no reason why M45 cannot be cracked out by this algorithm. In our quest as the aim of GIMPS I
request you dont dismiss this system lightly.

BTW: I am not impressed by your programming jargon and would have prefered an honest opinion with some worthwhile facts why this algortihm does not work and cannot be applied.

Remember Gimps is here to break records not cite the existing ones!

Mally :coffee:

[QUOTE=mfgoode;93150][QUOTE=bdodson;93093]Not to quibble, but you might have better luck if this thread
were moved to "math" or somewhere else more appropriate.
QUOTE]
:smile:
Thank you bdodson for the suggestion of redirecting my thread to the math forum. Typical beaucracy! Moderator please do the needful.

I am not sure what you mean by 'better luck'

I am just an amateur mathematician and by no means an all rounder. As such I have no knowledge of programming and have no desire of ever attaining it.
[/QUOTE]

I gave up my amateur status in 1976. You, OTOH don't seem to
fit the usual definition on amateur either. In particular, you've completely
missed the point of my comment. I don't program either. Runtime
analysis is part of computational number theory; independent of
coding.

[QUOTE]
Have you gone thru my worked examples to realise the potentiality of it?

I have merely given an algortihm and if you dont consider it as such lets call it a mathematical curiousity.

It is not the creation of a crank as I have resurrected this method by a respected Indian PhD who may have gone into obscurity had I not written about it. ( name witheld)

The reason why I have submitted it is to explore the possibility of working out a much shorter version as compared to the existing systems.
[/QUOTE]

No. I'm not immediately interested, for the reason explained in my
first note -- your description of the method shows no awareness of
the present state of research in primality testing. And besides, you're
posting in a "factoring" thread. The difference between the difficulty
of the problems should be clear to anyone with just a small amount
of attention. Those of us interested in factoring have just managed
to reach 200-decimal-digits (rsa-200), a record that replaced several
intermediate records, including RSA-155, at 512-bits. Professionals
in primality testing proved the primality of a 20,000-digit prime
("random", not mersenne or other special form), replacing their earlier
benchmark of 15,000 digits.

The difference is math, not coding. The complexity of the algorithm.
The term "polynomial -vs- subexponential" refers to the number of
computations needed to find the result.

[QUOTE]
I would add that if you can represent M44 then there is no reason why M45 cannot be cracked out by this algorithm. In our quest as the aim of GIMPS I
request you dont dismiss this system lightly.

BTW: I am not impressed by your programming jargon and would have prefered an honest opinion with some worthwhile facts why this algortihm does not work and cannot be applied.

Remember Gimps is here to break records not cite the existing ones!

Mally :coffee:[/QUOTE]

My honest opinion is that your posts are off-topic. If you scan through
the main mersenne web page (not this forum), you'll see that factoring;
in particular factoring Cunningham and related numbers; ... Ah. Maybe
not. We're listed under "Miscellaneous", then "links". The page I'm
attempting to refer to is

[url]http://www.mersenne.org/ecm.htm[/url]

With the exception of the Fermat numbers, all of the factoring questions
involve numbers with no more than 1200-bits, no more than
384-decimal-digits for the largest unfactored Cunningham. No 9-million
digit primes, we're looking for 70-digit (ecm) or perhaps 100-digit (gnfs)
primes. 'Cause our problem is more difficulty than proving primality, at
20000-digits (random primes) or 10-million-digits (mersenne).

You've wandered into a thread where the problems are different. And
the expertise needed to evaluate the algorithms are different also. I'm
requesting again that you wander elsewhere.

B. Dodson

[QUOTE=bdodson;93170][QUOTE=mfgoode;93150]

I gave up my amateur status in 1976. You, OTOH don't seem to
fit the usual definition on amateur either. In particular, you've completely
missed the point of my comment. I don't program either. Runtime
analysis is part of computational number theory; independent of
coding.
~ ~ ~
No. I'm not immediately interested, for the reason explained in my
first note -- You've wandered into a thread where the problems are different. And
the expertise needed to evaluate the algorithms are different also. I'm
requesting again that you wander elsewhere.

B. Dodson[/QUOTE]
:smile:
Thank you for your attention B. Dodson.
You are right and I have mistakenly wandered into a factoring thread when my algorithm does not deal with factoring, at most one or two divisions, to prove the primality or not of a number, beyond doubt.

I will take your advice and I request the moderator of this thread to please move it to the Math forum so I can open shop therein
Thank you once again,
Mally :coffee:

Mally,

If I understand correctly, you have stumbled onto what is known as a Fermat Probable Prime test, augmented with suggestions about how to perform the modulo that are well suited for hand calculation. Unfortunately, you have read Fermat's Little Theorem backwards, and you do not have a proof of primality: for any prime it is true that a^(p-1) = 1 mod p, but for every choice of "a" there are also some composites for which a^(n-1) = 1 mod n. Try googling PRP and/or "Probable Prime" for more information.

Also check out Strong Probable Prime or SPRP - this improves the PRP test by observing that as long as k is even, you can factor a^k-1 as a difference of squares, and that if p is prime, it must divide one of these. Many composites that return false positives to the PRP test will fail the SPRP test because n=a*b, but a divides one of the terms and b divides a different term, so n divides the product, but not any individual term.

[QUOTE=mfgoode;93082]:rolleyes:
Another worked example.

Since my last example was a bit laborious and confusing I give another example.

Let us test whether 524287 is a prime number or not. If it is then it must satisfy
2^(524287 -1) = 1 mod 524287.[/QUOTE]
According to your proof all Mersenne numbers with prime exponents are prime.

All congruences below are modulo [tex]2^p-1[/tex].

We trivially find that [tex]2^p \eq 1[/tex]

Now we will compute: [tex]2^{2^p-1} \,\pmod {2^p-1}[/tex]

Since all factors of [tex]2^p-1[/tex] have the form [tex]2kp+1[/tex] we know that [tex]2^p-1 \eq 1 \,\pmod p[/tex]

[tex]2^{2^p-1} \eq 2^{m*p+1} \eq 2*(2^p)^m \eq 2*1^m \eq 2[/tex]

So [tex]2^{2^p-1} \eq 2[/tex] or [tex]2^{2^p-2} \eq 1[/tex]

According to your criterion any Mersenne number is prime. :wink:

Right track
 

[QUOTE=wblipp;93224]Mally,

If I understand correctly, you have stumbled onto what is known as a Fermat Probable Prime test, augmented with suggestions about how to perform the modulo that are well suited for hand calculation. Unfortunately, you have read Fermat's Little Theorem backwards, and you do not have a proof of primality: for any prime it is true that a^(p-1) = 1 mod p, but for every choice of "a" there are also some composites for which a^(n-1) = 1 mod n. Try googling PRP and/or "Probable Prime" for more information.

Also check out Strong Probable Prime or SPRP - this improves the PRP test by observing that as long as k is even, you can factor a^k-1 as a difference of squares, and that if p is prime, it must divide one of these. Many composites that return false positives to the PRP test will fail the SPRP test because n=a*b, but a divides one of the terms and b divides a different term, so n divides the product, but not any individual term.[/QUOTE]

:bow:
I agree with what you say and thats a call to go back to the drawing board.
Until such time I will refrain from commenting on your post.
Thank you William, you have once again guided me on the right track
Mally :coffee:

Strange logic
 

[QUOTE=alpertron;93229]According to your proof all Mersenne numbers with prime exponents are prime.

According to your criterion any Mersenne number is prime. :wink:[/QUOTE]

:smile:

I'm afraid I cant agree with your first line statement and your conclusion.

Please refer to Wblipp's post as he is nearer to the truth and desribes more accurately what I presented.

Mally :coffee:

The idea is that all Mersenne numbers with prime exponents are pseudoprimes in base 2 using Fermat's method, so your method will fail for Mersenne numbers.

Mally,

You may want to study Dario's (alpertron's) post more closely. He was making the same point I made - that some composite numbers will pass the test you propose. He provides a large class of examples; he points out that every Mesenne number with prime exponent will pass your proposed test.

William

Misleading.
 

[QUOTE=alpertron;93436]The idea is that all Mersenne numbers with prime exponents are pseudoprimes in base 2 using Fermat's method, so your method will fail for Mersenne numbers.[/QUOTE]

[QUOTE=Wblipp]You may want to study Dario's (alpertron's) post more closely. He was making the same point I made - that some composite numbers will pass the test you propose. He provides a large class of examples; he points out that every Mesenne number with prime exponent will pass your proposed test.

William [/QUOTE]

:smile:
With due respect to both you two outanding math'cians I quote above

The last lines in both the above posts are contradictory.

Very misleading indeed!

Mally :coffee:

Mally, English is not my primary language so maybe I expressed wrong myself.

But the idea is that I showed that all Mersenne numbers with prime exponents pass your _primality_ test when we believe that almost all members of this set are _composite_ numbers.

[QUOTE=alpertron;93436]The idea is that all Mersenne numbers with prime exponents are pseudoprimes in base 2 using Fermat's method, so your method will fail for Mersenne numbers.[/QUOTE]

[QUOTE=wblipp;93447]he points out that every Mesenne number with prime exponent will pass your proposed test.[/QUOTE]


[QUOTE=mfgoode;93491]
The last lines in both the above posts are contradictory.[/QUOTE]

We seem to have a communications failure here; these two messages are attempting to say the same thing. Perhaps the problem lies in the use of the word "failure". Alpertron is saying that every Mersenne number with prime exponent (for example, 2[sup]11[/sup]-1 = 2047) will pass your proposed test and be declared prime by your proposed test. But 2047=23*89, so your proposed test is wrong when it declares 2047 to be prime. He calls this wrong answer a failure of your test.

I called this same event "passing your proposed test" because your proposed test says "prime" - but says so incorrectly.

It's a matter of focusing on the proposed test or the validity of the proposed test - we all agree the proposed test says 2047 is prime and we all agree the proposed test is wrong about 2047. And we've both been trying to say this.

My mistake.
 

:smile:
I'm very sorry that I misunderstood your answer Alpertron, but no matter, you are perfectly clear .

Thank you Wblipp for clarifying my misinterpretation of the posts.

I think you are over emphasising the question of a number passing the test as a pseudo prime or Poulet number. Once this method singles out a possible prime then the other pseudoprime tests can follow but it saves a lot of work.
If I recall correctly the Korsett's criterion can be applied for pseudoprimes.

And to my luck I have one last example left worked out for me that exactly shows that the very number Dario has picked viz: 2047 has been proved that it is NOT a prime by the same method I have presented altho' I made my examples as straightforward as I could. It simply does Not by pass the
algorithm and is detected

I would call this Divine intervention believe it or not.!

If 2047 is detected as a pseudoprime then by induction the rest of the primes should follow suit. Not rigorous, but all mathematicians particularly Fermat depended on intuition and hunches to achieve correct results.

So I cross my fingers and hope for the best. So far I have not personally tested this algorithm for random primes, myself ,to be honest but will now attempt to do so

I will definitely present this ingenious method of proving that 2047 is NOT prime using Fermat's theorem exatly as in my worked examples but I have a long journey ahead of me early morning so Please wait for it till I come back in the evening.

You may try this method out on 2047 but I warn you its a very tricky derivation if you have not , so far followed fully, my two worked examples.
Please get the gist of them, as if not, then you will not follow the one I present.

The originator has forethought this in the computer era and not meant for hand or calculator calculations. This man is a genious and hats off to him.

Mally :coffee:

[QUOTE=mfgoode;93512]You may try this method out on 2047 but I warn you its a very tricky derivation if you have not , so far followed fully, my two worked examples.[/QUOTE]

Are you planning on changing your method? As presented so far, you method would result in:

2[sup]2046[/sup]-1 = (2[sup]1023[/sup]-1)*(2[sup]1023[/sup]+1)
=(2[sup]11*93[/sup]-1)*(2[sup]11*93[/sup]+1)
=(2048[sup]93[/sup]-1)*(2048[sup]93[/sup]+1)
=((N+1)[sup]93[/sup]-1)*((N+1)[sup]93[/sup]+1)
=N*(N+2)
=N

Thereby falsely claiming 2047 is prime.

[QUOTE=wblipp;93514]Are you planning on changing your method? As presented so far, you method would result in:

2[sup]2046[/sup]-1 = (2[sup]1023[/sup]-1)*(2[sup]1023[/sup]+1)

Thereby falsely claiming 2047 is prime.[/QUOTE]
:smile:
No William, your reasoning is circular. It is like X = a^2 - b^2 so---X=X
I have no time right now but to shortly reply to you but will do so when I get back.
Rgds.

Mally :coffee:

Number 2047 Mersenne pseudoprime
 

:smile:
Number 2047: This number has factors 89 and 23. It is = 2^11 – 1 a Mersenne number and composite.

Let us test if this number is a prime or not. And whether it has factors or not by the application of FLT and Binomial theorems.

2 ^ (2047 – 1) – 1 = 2^2046 – 1
= (2^1023 + 1) (2^1023 – 1)

Now 2^1023 =[ 2^30 * 2] ^ 33 (Since 1023 = 31 * 33)
We have from our tables 2^30 = 1073741824

So 2^1023 = [1073741824 * 2]^33
Express 1073741824 in units of 2047

Thus 1073741824 = (524544) * (2047 ) + 256 (remainder)
If (524544)(2047) is denoted by N

Then 2^30 = N + 256 (N is 2047 Or its multiple )
Therefore 2^31 =2 N + 512 (Here 2N is a multiple so call it N )

Therefore 2^1023 = (N + 512)^33
= ( N + 511 and 3/4 + 1/4 )^33 But 511and ¾ = 2047/4

So= (5N/4 + ¼)^33

Now 2^1023 + 1 = (5N/4 +1/4)^33 + 1

2^1023 – 1 = (5N/4 + ¼) ^33 - 1

2^1023 = (5N/4 + ¼)^33

= ¼^33 (5N + 1)^33 (taking out the quarter)
Squaring both sides and subtracting 1 we get2^2046 – 1 = (1/4)^66 (5N + 1)^66 – 1

The last term of the expansion ( ¼)^ 66 is less numerically than – 1.

Therefore the expansion is not fully divisible.

Therefore by FLT 2047 is not a prime

Since 2047 is a Mersenne pseudoprime I presume that all mersenne pseduoprimes can be detected by this method and will not pass thru the prime sieve where the primes are let thru.

However a number like 341 an FLT pseudoprime will not be detected thru the sieve

A good test would be to see whether 2^67 can be detected or not by this algorithm. I have not attempted but leave it for you to test it out .

Mally :coffee:

[QUOTE=mfgoode;93697]The last term of the expansion ( ¼)^ 66 is less numerically than – 1.

Therefore the expansion is not fully divisible.

Therefore by FLT 2047 is not a prime[/QUOTE]

BOOM

That was the sound of my head asploding :rolleyes:

Mally, are you claiming that 2 ^ (2047 – 1) – 1 is not divisible by 2047?!

[QUOTE=mfgoode;93697]Now 2^1023 =[ 2^30 * 2] ^ 33 (Since 1023 = 31 * 33)[/QUOTE]

This works for a little number but not for a big one where you will no be able to factorise half of your number easily (1023 coming from 2047)

One thing is certain to me : for a number like 2047 this method is a bit ... complicated ?

[QUOTE=axn1;93698]Mally, are you claiming that 2 ^ (2047 – 1) – 1 is not divisible by 2047?![/QUOTE]

I think he is claiming that 2[sup]1023[/sup]-1 is not divisible by 2047.

But that''s wrong, too. I haven't taken the time figure out where, though.

Misinterprtation of calculation
 

[QUOTE=wblipp;93716]I think he is claiming that 2[sup]1023[/sup]-1 is not divisible by 2047.

But that''s wrong, too. I haven't taken the time figure out where, though.[/QUOTE]
:smile:

[Quote=axn1]Mally, are you claiming that 2 ^ (2047 – 1) – 1 is not divisible by 2047?![/QUOTE]

Yes. Try it out fellah on your p.c. if you like. Thats the beauty of the algorithm.

William you forgot the last line

[Quote= Mally]Squaring both sides and subtracting 1 we get2^2046 – 1 = (1/4)^66 (5N + 1)^66 – 1][/QUOTE]

I admit it is complicated no doubt but now I will try out 67 which is also a mersenne pseudoprime. I have the factors of 2^67 which I can present any time

Thanks for all the replies and Jacob it is, as you say, but involves at most one division of a known number which is 2^30.

In haste;

Mally :coffee:

[QUOTE=axn1;93698]
Mally, are you claiming that 2 ^ (2047 – 1) – 1 is not divisible by 2047?![/QUOTE]
[QUOTE=mfgoode;93728]
Yes. Try it out fellah on your p.c. if you like. Thats the beauty of the algorithm.
[/QUOTE]

My PC says that 2047 is a divisor of 2[sup]2046[/sup]-1. Anybody can check this on alpertron's Java factoring applet at

[url]http://www.alpertron.com.ar/ECM.HTM[/url]

copy and paste (2^2046-1)/2047

When the division is not exact, the applet reports a non-integer division.

Or better, because it looks up the divisors of full Mersenne numbers, just put in 2^2046-1 and observe that both 23 and 89 show up.

Mally - does this divisibility change your stance on the validity of your proof?

[QUOTE=mfgoode;93697]
Squaring both sides and subtracting 1 we get2^2046 – 1 = (1/4)^66 (5N + 1)^66 – 1

The last term of the expansion ( ¼)^ 66 is less numerically than – 1.

Therefore the expansion is not fully divisible.[/QUOTE]

Here is the mistake. Explicated more fully, the logic is that after binomial expansion we have 67 terms, and the first 66 are integers, but the last term is not an integer, so the entire expression is not an integer. The error is the unstated assumption that the first 66 terms are integers.

[QUOTE=wblipp;93716]I think he is claiming that 2[sup]1023[/sup]-1 is not divisible by 2047.

But that''s wrong, too. I haven't taken the time figure out where, though.[/QUOTE]
Of course it is wrong : 2[sup]1023[/sup]-1 is divisble by 2047 because 11 divides 1023 and that 2[sup]11[/sup]=2048=2047+1 and (a+1)[sup]n[/sup] leaves a reminder of 1 when divided by a if a and n are integers.

Mistake located.
 

[QUOTE=wblipp;93730]My PC says that 2047 is a divisor of 2[sup]2046[/sup]-1. Anybody can check this on alpertron's Java factoring applet at

[url]http://www.alpertron.com.ar/ECM.HTM[/url]

copy and paste (2^2046-1)/2047

When the division is not exact, the applet reports a non-integer division.

Or better, because it looks up the divisors of full Mersenne numbers, just put in 2^2046-1 and observe that both 23 and 89 show up.

Mally - does this divisibility change your stance on the validity of your proof?[/QUOTE]
:smile:
Thank you William and all the others for pointing out a flaw. I was wondering what was wrong in the calculation as I tried Dario's applet myself. What a remarkable device indeed Hats of to you alpertron. There is no doubt now that 2047 indeed divides the expression. Hence since it is composite and satisfies the FLT it is a pseudoprime.

But what went wrong with the calculation? This morning I got the answer.
The remainder 256 has been rounded of. It is actually 256 decimal and that makes a world of a difference.

No I dont change my stance on the proof. Mathematically it is correct.
I'm trying to devise a method on the same lines that shows that 2047 though composite passes the test and is indeed a pseudoprime. Unless 2047 is factorised it can not be certain that it is a prime or not but a pseudoprime.

I will try some actual small primes to test the algorithm and let you know my findings
Regards,

Mally :coffee:

Disregard.
 

[QUOTE=mfgoode;93959]:smile:

But what went wrong with the calculation? This morning I got the answer.
The remainder 256 has been rounded of. It is actually 256 decimal and that makes a world of a difference.

Regards,
Mally :coffee:[/QUOTE]

:sad:
Kindly disregard my post immediatey above.
The remiander 256 is correct and not ending in a fraction. How can it be when the two divisors are integers?
How ever your points lead me to another interpretation of the division which may prove useful and better than the original theory.

Mally :coffee: