- View Source--- In primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

> Here is an example with a 42nd root of unity:

More impressively, here is one with a 3414th root of unity:

p = 43334121400711;

a = 483020123189;

print(znorder(Mod(a,p)));

3414

It fools these tests:

> 1. p is an odd positive integer,

but not my gremlin-trap:

> 2. a is an integer with kronecker(a,p) = -1,

> 3. a^((p-1)/2) = -1 mod p,

> 5. Mod(1+x,x^2-a)^p = 1-x mod p,

> 4. gcd(a^6-1,p) = 1,

since the gcd extracts a factor:

print(gcd(a^6-1,p));

384634897

Puzzle: Find a pseudoprime, p,

with a false witness, a,

that fools (1), (2), (3), (5) and has

znorder(Mod(a,p)) > 3414.

Comment: I think it unlikely that you will also fool (4).

David - View Source--- In primenumbers@yahoogroups.com,

"bhelmes_1" <bhelmes@...> wrote:

> 1. let a jacobi (a, p)=-1

[4] is meaningless, as it stands.

> 2. let a^(p-1)/2 = -1 mod p

> 3. a^6 =/= 1 mod p

> 4. (1+sqrt (a))^p = 1-sqrt (a)

You should write a double mod:

4. (1+x)^p = 1-x mod(x^2-a,p)

> 1. Is it possible that there are other exceptions

There is no reason whatsoever to believe that

[1] to [4] establish that p is prime. Morevoer,

some folk believe that, for every epsilon > 0,

the number of pseudoprimes less than x may

exceed x^(1-epsilon), for /sufficiently/ large x.

> 2....

The group of units (Z/nZ)* is /not/ cyclic

> there is a cyclic order ...

> 3....

> there is a cyclic order ...

if n has at least two distinct odd prime fators.

David