## Re: [PrimeNumbers] Prime Test polynomial

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• I do agree that ... but I do not agree that ... For example, one root of f(x) is r1 = 0.64879067514879204822278147415791168556... and f( r^2 ) =
Message 1 of 6 , Nov 4 8:09 PM
I do agree that

> f(x**2) = 2 x**4 + 2 x**3 -4 (mod x**5 - x**3 - 2x**2 + 1)

but I do not agree that

> 2 x**4 + 2 x**3 -4 = 0 mod 2
> if x is ANY root of
> f(x) = x**5 - x**3 - 2x**2 + 1,

For example, one root of f(x) is
r1 = 0.64879067514879204822278147415791168556...

and
f( r^2 ) = 0.584270441039927278546515102139504666...

which is not zero mod 2.

Note that the roots of f are not integers, and therefore
2 x is not = 0 (mod 2), i.e. a multiple of 2.

Maximilian
• ... I see. Thank you for pointing out this distinction. It is only in the ring polynomials mod ( x**5 - x**3 - 2x**2 + 1, 2) that (x**2)**5 - (x**2)**3 -
Message 2 of 6 , Nov 5 8:43 AM
On 11/4/2010 11:09 PM, Maximilian Hasler wrote:
>
> I do agree that
>
>> f(x**2) = 2 x**4 + 2 x**3 -4 (mod x**5 - x**3 - 2x**2 + 1)
>
> but I do not agree that
>
>> 2 x**4 + 2 x**3 -4 = 0 mod 2
>> if x is ANY root of
>> f(x) = x**5 - x**3 - 2x**2 + 1,
>
> For example, one root of f(x) is
> r1 = 0.64879067514879204822278147415791168556...
>
> and
> f( r^2 ) = 0.584270441039927278546515102139504666...
>
> which is not zero mod 2.
>
> Note that the roots of f are not integers, and therefore
> 2 x is not = 0 (mod 2), i.e. a multiple of 2.

I see.

Thank you for pointing out this distinction.

It is only in the ring

polynomials mod ( x**5 - x**3 - 2x**2 + 1, 2)

that (x**2)**5 - (x**2)**3 - 2(x**2)**2 + 1 = 0

Is it true that for every positive prime p,

that in the ring

polynomials mod ( x**5 - x**3 - 2x**2 + 1, p),

that (x**p)**5 - (x**p)**3 - 2 (x**p)**2 + 1 = 0 ?

Is this type of relationship true for most irreducible polynomials?

for all irreducible polynomials?

for any polynomial?

What condition, if any, must be imposed on a polynomial F,
in order for it to be true,

that for any positive prime q,

that in the ring of polynomials mod (F(x),q)

that F(x**q) = 0?

Kermit
• ... Yes. Moreover there are no pseudoprimes less than 10^5: f(x)=x^5-x^3-2*x^2+1; for(p=2,10^5,if(isprime(p)!=(0==f(Mod(Mod(1,p)*x,f(x))^p)),print(p))); [The
Message 3 of 6 , Nov 5 9:18 AM

> Is it true that for every positive prime p,
> that in the ring polynomials mod ( x**5 - x**3 - 2x**2 + 1, p),
> that (x**p)**5 - (x**p)**3 - 2 (x**p)**2 + 1 = 0 ?

Yes. Moreover there are no pseudoprimes less than 10^5:

f(x)=x^5-x^3-2*x^2+1;
for(p=2,10^5,if(isprime(p)!=(0==f(Mod(Mod(1,p)*x,f(x))^p)),print(p)));

[The rest is silence.]

David
• ... Yes. I am ok with using the language of ring polynomial mod ( particular polynomial, integer). I will not quibble with you about whether or not it is
Message 4 of 6 , Nov 5 11:54 AM
On 11/5/2010 2:22 PM, Maximilian Hasler wrote:
>
>>
>>
>>
>>> Is it true that for every positive prime p,
>>> that in the ring polynomials mod ( x**5 - x**3 - 2x**2 + 1, p),
>>> that (x**p)**5 - (x**p)**3 - 2 (x**p)**2 + 1 = 0 ?
>
> This is a valid formulation ; not so, to my eyes, the one in terms of
> the roots of that polynomial, which are irrational numbers x=r[i] for
> which e.g. p*x is not an element of pZ<=> congruent to zero (mod p).
>
> M.
>

Yes. I am ok with using the language of

ring polynomial mod ( particular polynomial, integer).

I will not quibble with you about whether or not it is valid to
define parity for numbers in algebraic extensions of integers mod 2.

Now that that is settled, how about my question?

>>> Is it true that for every positive prime p,
>>> that in the ring polynomials mod ( x**5 - x**3 - 2x**2 + 1, p),
>>> that (x**p)**5 - (x**p)**3 - 2 (x**p)**2 + 1 = 0 ?

Kermit
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