- Assuming N odd, First counter-examples with their prime factors :

1729 { 7, 13, 19}

2821 { 7, 13, 31}

7921 { 89}

8911 { 7, 19, 67}

13699 { 7, 19, 103}

15841 { 7, 31, 73}

18721 { 97, 193}

29341 { 13, 37, 61}

31621 { 103, 307}

32761 { 181}

41041 { 7, 11, 13, 41}

46657 { 13, 37, 97}

49141 { 157, 313}

52633 { 7, 73, 103}

63973 { 7, 13, 19, 37}

64177 { 29, 2213}

75361 { 11, 13, 17, 31}

83333 { 167, 499}

93961 { 7, 31, 433}

115921 { 13, 37, 241}

126217 { 7, 13, 19, 73}

129481 { 11, 79, 149}

162401 { 17, 41, 233}

.......

if( N < 10 000 000) whe have 123 failures.

Regards,

JT

>the following program tries to determine the primality of a number by

10 cls

>combining two tests into one, avoiding the usual pseudo-prime situ-

>ation:

20 input N

30 NN= int((N-1)/2)

40 X= (2^N-3^NN-(N-1)) mod N

50 Y = N mod 5

60 print N;X

70 if (X=2 or X=4) and Y<>0 then print "prime"

80 goto 10

>it inevitably misses the 2, 3, and 5, but catches all the rest as far

<find a flaw or counter-example in this unproven technique ???

>as I can test it.

>my computer is somewhat outdated, but it told me that there were 168

>primes before the number 1001, when I used this code inside a for-

>loop.

>i am aware of the law of small numbers, but can anyone realistically

>Bill B

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http://www.echolalie.com

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[Non-text portions of this message have been removed] - Hello Bill,

> the following program tries to determine the primality of a number by

Slightly less programatic description of the test: If both

> combining two tests into one, avoiding the usual pseudo-prime situ-

> ation:

1) N must not be divisible by 5, and

2) 2^N - 3^((N-1)/2) = 1 or 3 (mod N),

then N is declared prime.

Clearly, this test is surely not stronger (although it is likely strictly

weaker) than a combination of

1) Divisibility by 5 test,

2) Base 2 Fermat test and

3) Base 3 Euler test

while having about the same time-complexity as these three together (in

fact, performing the step 3 only if step 2 succeeds can improve the

actual speed).

Combining the tests the way you did is usually not a good idea, as you're

losing some potentially useful information. In particular, instead of

checking 2^N = 2 (mod N) and 3^((N-1)/2) = +/-1 mod N, you're just

checking if their difference is equal to something -- which can, unless

proven otherwise, happen a lot more often.

Peter

--

[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278