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Prime Test polynomial

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• http://www.mathpages.com/home/kmath347/kmath347.htm claims that the first exception to the conjecture that n is a prime if and only if f(r**n) = 0 (mod n)
Message 1 of 6 , Nov 3, 2010
http://www.mathpages.com/home/kmath347/kmath347.htm

claims that the first exception to the conjecture that

n is a prime if and only if f(r**n) = 0 (mod n)
where r is any root of f(x) = x**5 - x**3 - 2x**2 + 1.

is n = 2258745004684033 = 27439297 * 82317889.

What is x**n mod n if

x**5 = x**3 + 2 x**2 - 1?

what procedure would be used to calculate

the roots x**n mod n

when x is constrained to satisfy

x**5 = x**3 + 2 x**2 - 1?

x**6 = x**4 + 2 x**3 - x

x**7 = x**5 + 2 x**4 - x**2

x**7 = (x**3 + 2 x**2 - 1) + 2 x**4 - x**2

x**7 = 2 x**4 + x**3 + (2 - 1) x**2 - 1

x**8 = 2 x**5 + x**4 + (2 - 1) x**3 - x

x**8 = 2 [x**3 + 2 x**2 - 1] + x**4 + (2-1) x**3 - x

x**8 = x**4 + (2+ 2 - 1) x**3 + (2 * 2) x**2 - x - 2*1

Clearly this implies a recurrence relation on the
coefficients of x**4, x**3, x**2, x, and 1.

Then it is clear that

x**n = 0 mod n if and only if

x**n = a4 x**4 + a3 x**3 + a2 x**2 + a1 x + a0
is, in mod n, some polynomial multiple of

x**3 + 2 x**2 - 1 .

2 x**4 + x**3 + (2 - 1) x**2 - 1

2 x**4 + x**3 + x**2 - 1

(x**3+2 x**2 -1)(b4 x**4 +b3 x**3 +b2 x**2 +b1 x + b0) =0 mod 7

reduces to a 4th degree equation = 0 mod 7,
which supposedly determines unique values of
b4, b3, b2, b1, b0.

It is reasonable to guess that if n is prime,
that the corresponding b4,b3,b2,b1,b0 can always
be solved mod n,

but if n is not prime, it might not be possible
to solve for the b4,b3,b2,b1,b0.

It is still not clear what algorithm had probably been used to prove
that x**n is not 0 mod n for each composite number tested.

Another approach:

f(x) = x**5 - x**3 - 2x**2 + 1.

f(x**2) = x**10 - x**6 - 2 x**4 + 1
= x**10 - x**6 + 1 mod 2

f(x**3) = x**15 - x**9 - 2 x**6 + 1
= x**15 - x**9 - 2 x**6 + 1 mod 3

Is this polynomial = 0 mod 3?

supposedly we could use the recurrence relations to reduce
f(x**m) to a fourth degree polynomial, and determine whether or not it
is equivalent to 0 mod m.

Does anyone see a simpler approach?

Kermit
• ... f(x) = x**5 - x**3 - 2x**2 + 1. f(r**n) = 0 (mod n) f(x**2) = x**10 - x**6 - 2 x**4 + 1 if f(x) = 0, then x**5 = x**3 + 2 x**2 - 1 x**6 = x**4 + 2 x**3 - x
Message 2 of 6 , Nov 4, 2010
On 11/4/2010 3:54 PM, Maximilian Hasler wrote:
>
> er, I'm not sure to understand.
>
> For primes p=2 or p=3, do we have f(r^p) = 0 (mod p) ??
>
> If so, for which of the roots ?
>
> M.

f(x) = x**5 - x**3 - 2x**2 + 1.

f(r**n) = 0 (mod n)

f(x**2) = x**10 - x**6 - 2 x**4 + 1

if f(x) = 0, then

x**5 = x**3 + 2 x**2 - 1
x**6 = x**4 + 2 x**3 - x
x**7 = x**5 + 2 x**4 - x**2
x**8 = x**6 + 2 x**5 - x**3
x**9 = x**7 + 2 x**6 - x**4
x**10 = x**8 + 2 x**7 - x**5

f(x**2) = x**10 - x**6 - 2 x**4 + 1
= [x**8 + 2 x**7 - x**5] - [x**4 + 2 x**3 - x ] - 2 x**4 + 1

= x**8 + 2 x**7 - x**5 - x**4 - 2 x**3 + x - 2 x**4 + 1
= x**8 + 2 x**7 - x**5 - 3 x**4 - 2 x**3 + x - 1
= [x**6 + 2 x**5 - x**3] + 2 x**7 - x**5 - 3 x**4 - 2 x**3 + x - 1
= 2 x**7 + x**6 + x**5 - 3 x**4 - 3 x**3 + x - 1
= 2 [x**5 + 2 x**4 - x**2] + x**6 + x**5 - 3 x**4 - 3 x**3 + x - 1
= x**6 + 3 x**5 + x**4 - 3 x**3 - 2 x**2 + x - 1
= [ x**4 + 2 x**3 - x ] + 3 x**5 + x**4 - 3 x**3 - 2 x**2 + x - 1
= 3 x**5 + 2 x**4 - x**3 - 2 x**2 - 1
= 3 [ x**3 + 2 x**2 - 1 ] + 2 x**4 - x**3 - 2 x**2 - 1
= 2 x**4 + 2 x**3 -4 = 0 mod 2

Proves that for the prime 2, if x is ANY root of

f(x) = x**5 - x**3 - 2x**2 + 1,

then

f(x**2) = 0 mod 2.

Similar laborious proof can be constructed for
f(x**3) = 0.

Kermit
• 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 3 of 6 , Nov 4, 2010
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 4 of 6 , Nov 5, 2010
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 5 of 6 , Nov 5, 2010
Kermit Rose <kermit@...> asked:

> 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 6 of 6 , Nov 5, 2010
On 11/5/2010 2:22 PM, Maximilian Hasler wrote:
>
> On Fri, Nov 5, 2010 at 10:18 AM, djbroadhurst<d.broadhurst@...> wrote:
>>
>>
>> --- In primenumbers@yahoogroups.com,
>> Kermit Rose<kermit@...> asked:
>>
>>> 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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