Discrete Logarithm Problem in Finite Fields












2














Consider the finite field $mathbb{F}_{displaystyle{3^{23}}}$ which is constructed from the primitive polynomial ${bf f}=x^{23}-x^3+1$ over $mathbb{F}_3$. Let $alpha$ be a root of $bf f$. Suppose that $beta$ is a polynomial based on the $alpha$ such as $beta=alpha^{22}+alpha^{21}+alpha^{11}+1
$.



My question: Is there a method (except full search)
to find a positive integer number $k$ such that
$beta=alpha^{k}$?



I know my question is a kind of discrete logarithm problem, but I have no idea except exhaustive search to answer.



Thanks for any suggestions






finite-fields







share|cite|improve this question


















  • 1




    Index calculus may work. Here is an example run, though in a much smaller field. I'm fairly sure I don't want to try that in your case (without a serious silicon aide) unless the polynomial factors easily.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:12








  • 1




    $3^{23}-1=2cdot47cdot1001523179$, so Chinese remainder theorem is not going to really help. Baby step - giant step will give you square root order reduction. Pollard $rho$ about the same.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:15






  • 1




    Index calculus has a wikipedia page. See that for a general description and links to more. I jotted down that example partly to have a semiserious case as an example I can refer to.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:28










  • @JyrkiLahtonen At first, I want to thank you professor Lahtonen because of useful link and notations that you mentioned. I have learned from you that finite fields may seem hard, but with some sweet tricks. Because of this, I want to ask my question clearly. Consider $gamma$ is an element in $mathbb{F}_{displaystyle{3^{23}}}$ such that its order is $47$. Now, consider the following polynomial $$ beta=gamma^{22}+gamma^{21}+gamma^{19}+gamma^{18}+gamma^{15}-gamma^{14}-gamma^{13}+gamma^{10}-gamma^{9}-gamma^{8}-gamma^{7}+gamma^{6}-gamma^{3}- gamma+1 $$
    – user0410
    Dec 27 '18 at 12:18






  • 1




    $gamma$ generates that field, so every element of $Bbb{F}_{3^{23}}$ can be written as a polynomial in $gamma$. Also, this expression does not specify $beta$ uniquely because there are $46$ primitive roots of unity of order $47$, split into two conjugacy classes. This calculation may be one for the computers. I'm afraid I don't see tricks for handling such a $beta$.
    – Jyrki Lahtonen
    Dec 27 '18 at 16:58
















2














Consider the finite field $mathbb{F}_{displaystyle{3^{23}}}$ which is constructed from the primitive polynomial ${bf f}=x^{23}-x^3+1$ over $mathbb{F}_3$. Let $alpha$ be a root of $bf f$. Suppose that $beta$ is a polynomial based on the $alpha$ such as $beta=alpha^{22}+alpha^{21}+alpha^{11}+1
$.



My question: Is there a method (except full search)
to find a positive integer number $k$ such that
$beta=alpha^{k}$?



I know my question is a kind of discrete logarithm problem, but I have no idea except exhaustive search to answer.



Thanks for any suggestions






finite-fields







share|cite|improve this question


















  • 1




    Index calculus may work. Here is an example run, though in a much smaller field. I'm fairly sure I don't want to try that in your case (without a serious silicon aide) unless the polynomial factors easily.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:12








  • 1




    $3^{23}-1=2cdot47cdot1001523179$, so Chinese remainder theorem is not going to really help. Baby step - giant step will give you square root order reduction. Pollard $rho$ about the same.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:15






  • 1




    Index calculus has a wikipedia page. See that for a general description and links to more. I jotted down that example partly to have a semiserious case as an example I can refer to.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:28










  • @JyrkiLahtonen At first, I want to thank you professor Lahtonen because of useful link and notations that you mentioned. I have learned from you that finite fields may seem hard, but with some sweet tricks. Because of this, I want to ask my question clearly. Consider $gamma$ is an element in $mathbb{F}_{displaystyle{3^{23}}}$ such that its order is $47$. Now, consider the following polynomial $$ beta=gamma^{22}+gamma^{21}+gamma^{19}+gamma^{18}+gamma^{15}-gamma^{14}-gamma^{13}+gamma^{10}-gamma^{9}-gamma^{8}-gamma^{7}+gamma^{6}-gamma^{3}- gamma+1 $$
    – user0410
    Dec 27 '18 at 12:18






  • 1




    $gamma$ generates that field, so every element of $Bbb{F}_{3^{23}}$ can be written as a polynomial in $gamma$. Also, this expression does not specify $beta$ uniquely because there are $46$ primitive roots of unity of order $47$, split into two conjugacy classes. This calculation may be one for the computers. I'm afraid I don't see tricks for handling such a $beta$.
    – Jyrki Lahtonen
    Dec 27 '18 at 16:58














2












2








2







Consider the finite field $mathbb{F}_{displaystyle{3^{23}}}$ which is constructed from the primitive polynomial ${bf f}=x^{23}-x^3+1$ over $mathbb{F}_3$. Let $alpha$ be a root of $bf f$. Suppose that $beta$ is a polynomial based on the $alpha$ such as $beta=alpha^{22}+alpha^{21}+alpha^{11}+1
$.



My question: Is there a method (except full search)
to find a positive integer number $k$ such that
$beta=alpha^{k}$?



I know my question is a kind of discrete logarithm problem, but I have no idea except exhaustive search to answer.



Thanks for any suggestions






finite-fields







share|cite|improve this question













Consider the finite field $mathbb{F}_{displaystyle{3^{23}}}$ which is constructed from the primitive polynomial ${bf f}=x^{23}-x^3+1$ over $mathbb{F}_3$. Let $alpha$ be a root of $bf f$. Suppose that $beta$ is a polynomial based on the $alpha$ such as $beta=alpha^{22}+alpha^{21}+alpha^{11}+1
$.



My question: Is there a method (except full search)
to find a positive integer number $k$ such that
$beta=alpha^{k}$?



I know my question is a kind of discrete logarithm problem, but I have no idea except exhaustive search to answer.



Thanks for any suggestions






finite-fields




finite-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 '18 at 22:18









user0410

1899




1899








  • 1




    Index calculus may work. Here is an example run, though in a much smaller field. I'm fairly sure I don't want to try that in your case (without a serious silicon aide) unless the polynomial factors easily.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:12








  • 1




    $3^{23}-1=2cdot47cdot1001523179$, so Chinese remainder theorem is not going to really help. Baby step - giant step will give you square root order reduction. Pollard $rho$ about the same.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:15






  • 1




    Index calculus has a wikipedia page. See that for a general description and links to more. I jotted down that example partly to have a semiserious case as an example I can refer to.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:28










  • @JyrkiLahtonen At first, I want to thank you professor Lahtonen because of useful link and notations that you mentioned. I have learned from you that finite fields may seem hard, but with some sweet tricks. Because of this, I want to ask my question clearly. Consider $gamma$ is an element in $mathbb{F}_{displaystyle{3^{23}}}$ such that its order is $47$. Now, consider the following polynomial $$ beta=gamma^{22}+gamma^{21}+gamma^{19}+gamma^{18}+gamma^{15}-gamma^{14}-gamma^{13}+gamma^{10}-gamma^{9}-gamma^{8}-gamma^{7}+gamma^{6}-gamma^{3}- gamma+1 $$
    – user0410
    Dec 27 '18 at 12:18






  • 1




    $gamma$ generates that field, so every element of $Bbb{F}_{3^{23}}$ can be written as a polynomial in $gamma$. Also, this expression does not specify $beta$ uniquely because there are $46$ primitive roots of unity of order $47$, split into two conjugacy classes. This calculation may be one for the computers. I'm afraid I don't see tricks for handling such a $beta$.
    – Jyrki Lahtonen
    Dec 27 '18 at 16:58














  • 1




    Index calculus may work. Here is an example run, though in a much smaller field. I'm fairly sure I don't want to try that in your case (without a serious silicon aide) unless the polynomial factors easily.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:12








  • 1




    $3^{23}-1=2cdot47cdot1001523179$, so Chinese remainder theorem is not going to really help. Baby step - giant step will give you square root order reduction. Pollard $rho$ about the same.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:15






  • 1




    Index calculus has a wikipedia page. See that for a general description and links to more. I jotted down that example partly to have a semiserious case as an example I can refer to.
    – Jyrki Lahtonen
    Dec 27 '18 at 6:28










  • @JyrkiLahtonen At first, I want to thank you professor Lahtonen because of useful link and notations that you mentioned. I have learned from you that finite fields may seem hard, but with some sweet tricks. Because of this, I want to ask my question clearly. Consider $gamma$ is an element in $mathbb{F}_{displaystyle{3^{23}}}$ such that its order is $47$. Now, consider the following polynomial $$ beta=gamma^{22}+gamma^{21}+gamma^{19}+gamma^{18}+gamma^{15}-gamma^{14}-gamma^{13}+gamma^{10}-gamma^{9}-gamma^{8}-gamma^{7}+gamma^{6}-gamma^{3}- gamma+1 $$
    – user0410
    Dec 27 '18 at 12:18






  • 1




    $gamma$ generates that field, so every element of $Bbb{F}_{3^{23}}$ can be written as a polynomial in $gamma$. Also, this expression does not specify $beta$ uniquely because there are $46$ primitive roots of unity of order $47$, split into two conjugacy classes. This calculation may be one for the computers. I'm afraid I don't see tricks for handling such a $beta$.
    – Jyrki Lahtonen
    Dec 27 '18 at 16:58








1




1




Index calculus may work. Here is an example run, though in a much smaller field. I'm fairly sure I don't want to try that in your case (without a serious silicon aide) unless the polynomial factors easily.
– Jyrki Lahtonen
Dec 27 '18 at 6:12






Index calculus may work. Here is an example run, though in a much smaller field. I'm fairly sure I don't want to try that in your case (without a serious silicon aide) unless the polynomial factors easily.
– Jyrki Lahtonen
Dec 27 '18 at 6:12






1




1




$3^{23}-1=2cdot47cdot1001523179$, so Chinese remainder theorem is not going to really help. Baby step - giant step will give you square root order reduction. Pollard $rho$ about the same.
– Jyrki Lahtonen
Dec 27 '18 at 6:15




$3^{23}-1=2cdot47cdot1001523179$, so Chinese remainder theorem is not going to really help. Baby step - giant step will give you square root order reduction. Pollard $rho$ about the same.
– Jyrki Lahtonen
Dec 27 '18 at 6:15




1




1




Index calculus has a wikipedia page. See that for a general description and links to more. I jotted down that example partly to have a semiserious case as an example I can refer to.
– Jyrki Lahtonen
Dec 27 '18 at 6:28




Index calculus has a wikipedia page. See that for a general description and links to more. I jotted down that example partly to have a semiserious case as an example I can refer to.
– Jyrki Lahtonen
Dec 27 '18 at 6:28












@JyrkiLahtonen At first, I want to thank you professor Lahtonen because of useful link and notations that you mentioned. I have learned from you that finite fields may seem hard, but with some sweet tricks. Because of this, I want to ask my question clearly. Consider $gamma$ is an element in $mathbb{F}_{displaystyle{3^{23}}}$ such that its order is $47$. Now, consider the following polynomial $$ beta=gamma^{22}+gamma^{21}+gamma^{19}+gamma^{18}+gamma^{15}-gamma^{14}-gamma^{13}+gamma^{10}-gamma^{9}-gamma^{8}-gamma^{7}+gamma^{6}-gamma^{3}- gamma+1 $$
– user0410
Dec 27 '18 at 12:18




@JyrkiLahtonen At first, I want to thank you professor Lahtonen because of useful link and notations that you mentioned. I have learned from you that finite fields may seem hard, but with some sweet tricks. Because of this, I want to ask my question clearly. Consider $gamma$ is an element in $mathbb{F}_{displaystyle{3^{23}}}$ such that its order is $47$. Now, consider the following polynomial $$ beta=gamma^{22}+gamma^{21}+gamma^{19}+gamma^{18}+gamma^{15}-gamma^{14}-gamma^{13}+gamma^{10}-gamma^{9}-gamma^{8}-gamma^{7}+gamma^{6}-gamma^{3}- gamma+1 $$
– user0410
Dec 27 '18 at 12:18




1




1




$gamma$ generates that field, so every element of $Bbb{F}_{3^{23}}$ can be written as a polynomial in $gamma$. Also, this expression does not specify $beta$ uniquely because there are $46$ primitive roots of unity of order $47$, split into two conjugacy classes. This calculation may be one for the computers. I'm afraid I don't see tricks for handling such a $beta$.
– Jyrki Lahtonen
Dec 27 '18 at 16:58




$gamma$ generates that field, so every element of $Bbb{F}_{3^{23}}$ can be written as a polynomial in $gamma$. Also, this expression does not specify $beta$ uniquely because there are $46$ primitive roots of unity of order $47$, split into two conjugacy classes. This calculation may be one for the computers. I'm afraid I don't see tricks for handling such a $beta$.
– Jyrki Lahtonen
Dec 27 '18 at 16:58















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