Primitive Trinomial for $82589933$?












6














At Twelve new primitive binary trinomials, $x^{74207281}+x^{9999621}+1$ is shown to be a primitive trinomial in $GF(2)[x]$.



Note that $2^{74207281}-1$ was the largest known (Mersenne) prime before 2018, that is the prime in the exponent of the Mersenne prime gave rise to a trinomial.



In the interim, new Mersenne primes have been found.



Just announced (Dec 2018), $2^{82589933}-1$ is prime, and in Jan 2018, $2^{77232917}-1$ was found to be prime.



Do these also give rise to trinomials?



To be explicit, are there $j$ or $k$ such that $x^{82589933}+x^{j}+1$ or $x^{77232917}+x^{k}+1$ are primitive trinomials in $GF(2)[x]$?






number-theory polynomials prime-numbers finite-fields







share|cite|improve this question




















  • 1




    I don't understand the votes to put this on hold. There was nothing unclear about the question. Anyway, when $p=2^m-1$ is a Mersenne prime, a polynomial $f(x)$ of degree $m$ is primitive in $Bbb{F}_2[x]$ if and only if it is irreducible if and only if $x^{2^m}equiv xpmod{f(x)}$. That last congruence can be checked with the good ole square-and-multipl if you can spare the CPU time.
    – Jyrki Lahtonen
    Dec 23 '18 at 4:04








  • 2




    Having said that, I'm not entirely sure that people here are well placed to answer this. I would guess that the search for such trinomials does take quite a bit of CPU resources even though testing any candidate is relatively straightforward.
    – Jyrki Lahtonen
    Dec 23 '18 at 4:13






  • 1




    I want to make it clear that the method I used here to find primitive trinomials of degree $127$ absolutely depended on the coincidence that $127$ is itself also a Mersenne prime. That trick doesn't work here.
    – Jyrki Lahtonen
    Dec 27 '18 at 17:09








  • 1




    I rearranged the question a bit; even the edited version, I found it hard to follow.
    – quid♦
    Dec 28 '18 at 13:49
















6














At Twelve new primitive binary trinomials, $x^{74207281}+x^{9999621}+1$ is shown to be a primitive trinomial in $GF(2)[x]$.



Note that $2^{74207281}-1$ was the largest known (Mersenne) prime before 2018, that is the prime in the exponent of the Mersenne prime gave rise to a trinomial.



In the interim, new Mersenne primes have been found.



Just announced (Dec 2018), $2^{82589933}-1$ is prime, and in Jan 2018, $2^{77232917}-1$ was found to be prime.



Do these also give rise to trinomials?



To be explicit, are there $j$ or $k$ such that $x^{82589933}+x^{j}+1$ or $x^{77232917}+x^{k}+1$ are primitive trinomials in $GF(2)[x]$?






number-theory polynomials prime-numbers finite-fields







share|cite|improve this question




















  • 1




    I don't understand the votes to put this on hold. There was nothing unclear about the question. Anyway, when $p=2^m-1$ is a Mersenne prime, a polynomial $f(x)$ of degree $m$ is primitive in $Bbb{F}_2[x]$ if and only if it is irreducible if and only if $x^{2^m}equiv xpmod{f(x)}$. That last congruence can be checked with the good ole square-and-multipl if you can spare the CPU time.
    – Jyrki Lahtonen
    Dec 23 '18 at 4:04








  • 2




    Having said that, I'm not entirely sure that people here are well placed to answer this. I would guess that the search for such trinomials does take quite a bit of CPU resources even though testing any candidate is relatively straightforward.
    – Jyrki Lahtonen
    Dec 23 '18 at 4:13






  • 1




    I want to make it clear that the method I used here to find primitive trinomials of degree $127$ absolutely depended on the coincidence that $127$ is itself also a Mersenne prime. That trick doesn't work here.
    – Jyrki Lahtonen
    Dec 27 '18 at 17:09








  • 1




    I rearranged the question a bit; even the edited version, I found it hard to follow.
    – quid♦
    Dec 28 '18 at 13:49














6












6








6


2





At Twelve new primitive binary trinomials, $x^{74207281}+x^{9999621}+1$ is shown to be a primitive trinomial in $GF(2)[x]$.



Note that $2^{74207281}-1$ was the largest known (Mersenne) prime before 2018, that is the prime in the exponent of the Mersenne prime gave rise to a trinomial.



In the interim, new Mersenne primes have been found.



Just announced (Dec 2018), $2^{82589933}-1$ is prime, and in Jan 2018, $2^{77232917}-1$ was found to be prime.



Do these also give rise to trinomials?



To be explicit, are there $j$ or $k$ such that $x^{82589933}+x^{j}+1$ or $x^{77232917}+x^{k}+1$ are primitive trinomials in $GF(2)[x]$?






number-theory polynomials prime-numbers finite-fields







share|cite|improve this question















At Twelve new primitive binary trinomials, $x^{74207281}+x^{9999621}+1$ is shown to be a primitive trinomial in $GF(2)[x]$.



Note that $2^{74207281}-1$ was the largest known (Mersenne) prime before 2018, that is the prime in the exponent of the Mersenne prime gave rise to a trinomial.



In the interim, new Mersenne primes have been found.



Just announced (Dec 2018), $2^{82589933}-1$ is prime, and in Jan 2018, $2^{77232917}-1$ was found to be prime.



Do these also give rise to trinomials?



To be explicit, are there $j$ or $k$ such that $x^{82589933}+x^{j}+1$ or $x^{77232917}+x^{k}+1$ are primitive trinomials in $GF(2)[x]$?






number-theory polynomials prime-numbers finite-fields




number-theory polynomials prime-numbers finite-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 13:47









quid♦

36.9k95093




36.9k95093










asked Dec 21 '18 at 21:31









Ed PeggEd Pegg

9,77232592




9,77232592








  • 1




    I don't understand the votes to put this on hold. There was nothing unclear about the question. Anyway, when $p=2^m-1$ is a Mersenne prime, a polynomial $f(x)$ of degree $m$ is primitive in $Bbb{F}_2[x]$ if and only if it is irreducible if and only if $x^{2^m}equiv xpmod{f(x)}$. That last congruence can be checked with the good ole square-and-multipl if you can spare the CPU time.
    – Jyrki Lahtonen
    Dec 23 '18 at 4:04








  • 2




    Having said that, I'm not entirely sure that people here are well placed to answer this. I would guess that the search for such trinomials does take quite a bit of CPU resources even though testing any candidate is relatively straightforward.
    – Jyrki Lahtonen
    Dec 23 '18 at 4:13






  • 1




    I want to make it clear that the method I used here to find primitive trinomials of degree $127$ absolutely depended on the coincidence that $127$ is itself also a Mersenne prime. That trick doesn't work here.
    – Jyrki Lahtonen
    Dec 27 '18 at 17:09








  • 1




    I rearranged the question a bit; even the edited version, I found it hard to follow.
    – quid♦
    Dec 28 '18 at 13:49














  • 1




    I don't understand the votes to put this on hold. There was nothing unclear about the question. Anyway, when $p=2^m-1$ is a Mersenne prime, a polynomial $f(x)$ of degree $m$ is primitive in $Bbb{F}_2[x]$ if and only if it is irreducible if and only if $x^{2^m}equiv xpmod{f(x)}$. That last congruence can be checked with the good ole square-and-multipl if you can spare the CPU time.
    – Jyrki Lahtonen
    Dec 23 '18 at 4:04








  • 2




    Having said that, I'm not entirely sure that people here are well placed to answer this. I would guess that the search for such trinomials does take quite a bit of CPU resources even though testing any candidate is relatively straightforward.
    – Jyrki Lahtonen
    Dec 23 '18 at 4:13






  • 1




    I want to make it clear that the method I used here to find primitive trinomials of degree $127$ absolutely depended on the coincidence that $127$ is itself also a Mersenne prime. That trick doesn't work here.
    – Jyrki Lahtonen
    Dec 27 '18 at 17:09








  • 1




    I rearranged the question a bit; even the edited version, I found it hard to follow.
    – quid♦
    Dec 28 '18 at 13:49








1




1




I don't understand the votes to put this on hold. There was nothing unclear about the question. Anyway, when $p=2^m-1$ is a Mersenne prime, a polynomial $f(x)$ of degree $m$ is primitive in $Bbb{F}_2[x]$ if and only if it is irreducible if and only if $x^{2^m}equiv xpmod{f(x)}$. That last congruence can be checked with the good ole square-and-multipl if you can spare the CPU time.
– Jyrki Lahtonen
Dec 23 '18 at 4:04






I don't understand the votes to put this on hold. There was nothing unclear about the question. Anyway, when $p=2^m-1$ is a Mersenne prime, a polynomial $f(x)$ of degree $m$ is primitive in $Bbb{F}_2[x]$ if and only if it is irreducible if and only if $x^{2^m}equiv xpmod{f(x)}$. That last congruence can be checked with the good ole square-and-multipl if you can spare the CPU time.
– Jyrki Lahtonen
Dec 23 '18 at 4:04






2




2




Having said that, I'm not entirely sure that people here are well placed to answer this. I would guess that the search for such trinomials does take quite a bit of CPU resources even though testing any candidate is relatively straightforward.
– Jyrki Lahtonen
Dec 23 '18 at 4:13




Having said that, I'm not entirely sure that people here are well placed to answer this. I would guess that the search for such trinomials does take quite a bit of CPU resources even though testing any candidate is relatively straightforward.
– Jyrki Lahtonen
Dec 23 '18 at 4:13




1




1




I want to make it clear that the method I used here to find primitive trinomials of degree $127$ absolutely depended on the coincidence that $127$ is itself also a Mersenne prime. That trick doesn't work here.
– Jyrki Lahtonen
Dec 27 '18 at 17:09






I want to make it clear that the method I used here to find primitive trinomials of degree $127$ absolutely depended on the coincidence that $127$ is itself also a Mersenne prime. That trick doesn't work here.
– Jyrki Lahtonen
Dec 27 '18 at 17:09






1




1




I rearranged the question a bit; even the edited version, I found it hard to follow.
– quid♦
Dec 28 '18 at 13:49




I rearranged the question a bit; even the edited version, I found it hard to follow.
– quid♦
Dec 28 '18 at 13:49










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