## Re: Prime Test polynomial

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• ... 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 1 of 6 , Nov 5, 2010

> 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 2 of 6 , Nov 5, 2010
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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