Calculations on $GF(16)$ find $0111/1111$












2












$begingroup$


It's my first time doing finite field arithmetics. As an exercise, I want to find $0111/1111 in GF(16)$ generated by $Pi(alpha)=1+alpha +alpha^4$ that is an irreducible polynomial.



In polynomial form we have:




  • $0111 rightarrow alpha+alpha^2+alpha^3$

  • $1111 rightarrow 1+alpha+alpha^2+alpha^3$


If I perform the polynomial division, I obtain $-1$ (that is the same result obtained writing $0111 equiv -1 pmod {1111}$).



How can I compute this result $-1$ in the right element of the field? Or perhaps this some kind of sign that the result $not in GF(16)$?






abstract-algebra finite-fields







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You basically compute the inverse of 1111 in the field and then multiply 0111 with the inverse of 1111.
    $endgroup$
    – Wuestenfux
    Dec 13 '18 at 11:35










  • $begingroup$
    So I need to calculate $1111^-1$, the problem is that this process in $GF(16)$ is not clear
    $endgroup$
    – Alessar
    Dec 13 '18 at 11:51






  • 1




    $begingroup$
    The extended euclidean algorithm gives a very clear way of doing it.
    $endgroup$
    – ancientmathematician
    Dec 13 '18 at 11:52






  • 1




    $begingroup$
    There is only one $GF(16)$, in the sense that all fields of 16 elements are isomorphic to each other. But there are many ways to represent an element of $GF(16)$, which depend on the specific irreducible polynomial used. So your question as it stands is badly specified. It is true that you mention $Pi(alpha)$, but only as an afterthought; whereas without it, we can't know what you mean by $0111$ and $1111$.
    $endgroup$
    – TonyK
    Dec 13 '18 at 12:34






  • 2




    $begingroup$
    When I said "many", I should in fact have said "three". In addition to your $Pi(alpha)$, there is $1+alpha^3+alpha^4$ and $1+alpha+alpha^2+alpha^3+alpha^4$.
    $endgroup$
    – TonyK
    Dec 13 '18 at 12:41
















2












$begingroup$


It's my first time doing finite field arithmetics. As an exercise, I want to find $0111/1111 in GF(16)$ generated by $Pi(alpha)=1+alpha +alpha^4$ that is an irreducible polynomial.



In polynomial form we have:




  • $0111 rightarrow alpha+alpha^2+alpha^3$

  • $1111 rightarrow 1+alpha+alpha^2+alpha^3$


If I perform the polynomial division, I obtain $-1$ (that is the same result obtained writing $0111 equiv -1 pmod {1111}$).



How can I compute this result $-1$ in the right element of the field? Or perhaps this some kind of sign that the result $not in GF(16)$?






abstract-algebra finite-fields







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You basically compute the inverse of 1111 in the field and then multiply 0111 with the inverse of 1111.
    $endgroup$
    – Wuestenfux
    Dec 13 '18 at 11:35










  • $begingroup$
    So I need to calculate $1111^-1$, the problem is that this process in $GF(16)$ is not clear
    $endgroup$
    – Alessar
    Dec 13 '18 at 11:51






  • 1




    $begingroup$
    The extended euclidean algorithm gives a very clear way of doing it.
    $endgroup$
    – ancientmathematician
    Dec 13 '18 at 11:52






  • 1




    $begingroup$
    There is only one $GF(16)$, in the sense that all fields of 16 elements are isomorphic to each other. But there are many ways to represent an element of $GF(16)$, which depend on the specific irreducible polynomial used. So your question as it stands is badly specified. It is true that you mention $Pi(alpha)$, but only as an afterthought; whereas without it, we can't know what you mean by $0111$ and $1111$.
    $endgroup$
    – TonyK
    Dec 13 '18 at 12:34






  • 2




    $begingroup$
    When I said "many", I should in fact have said "three". In addition to your $Pi(alpha)$, there is $1+alpha^3+alpha^4$ and $1+alpha+alpha^2+alpha^3+alpha^4$.
    $endgroup$
    – TonyK
    Dec 13 '18 at 12:41














2












2








2





$begingroup$


It's my first time doing finite field arithmetics. As an exercise, I want to find $0111/1111 in GF(16)$ generated by $Pi(alpha)=1+alpha +alpha^4$ that is an irreducible polynomial.



In polynomial form we have:




  • $0111 rightarrow alpha+alpha^2+alpha^3$

  • $1111 rightarrow 1+alpha+alpha^2+alpha^3$


If I perform the polynomial division, I obtain $-1$ (that is the same result obtained writing $0111 equiv -1 pmod {1111}$).



How can I compute this result $-1$ in the right element of the field? Or perhaps this some kind of sign that the result $not in GF(16)$?






abstract-algebra finite-fields







share|cite|improve this question











$endgroup$




It's my first time doing finite field arithmetics. As an exercise, I want to find $0111/1111 in GF(16)$ generated by $Pi(alpha)=1+alpha +alpha^4$ that is an irreducible polynomial.



In polynomial form we have:




  • $0111 rightarrow alpha+alpha^2+alpha^3$

  • $1111 rightarrow 1+alpha+alpha^2+alpha^3$


If I perform the polynomial division, I obtain $-1$ (that is the same result obtained writing $0111 equiv -1 pmod {1111}$).



How can I compute this result $-1$ in the right element of the field? Or perhaps this some kind of sign that the result $not in GF(16)$?






abstract-algebra finite-fields




abstract-algebra finite-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 12:38







Alessar

















asked Dec 13 '18 at 11:32









AlessarAlessar

308115




308115








  • 1




    $begingroup$
    You basically compute the inverse of 1111 in the field and then multiply 0111 with the inverse of 1111.
    $endgroup$
    – Wuestenfux
    Dec 13 '18 at 11:35










  • $begingroup$
    So I need to calculate $1111^-1$, the problem is that this process in $GF(16)$ is not clear
    $endgroup$
    – Alessar
    Dec 13 '18 at 11:51






  • 1




    $begingroup$
    The extended euclidean algorithm gives a very clear way of doing it.
    $endgroup$
    – ancientmathematician
    Dec 13 '18 at 11:52






  • 1




    $begingroup$
    There is only one $GF(16)$, in the sense that all fields of 16 elements are isomorphic to each other. But there are many ways to represent an element of $GF(16)$, which depend on the specific irreducible polynomial used. So your question as it stands is badly specified. It is true that you mention $Pi(alpha)$, but only as an afterthought; whereas without it, we can't know what you mean by $0111$ and $1111$.
    $endgroup$
    – TonyK
    Dec 13 '18 at 12:34






  • 2




    $begingroup$
    When I said "many", I should in fact have said "three". In addition to your $Pi(alpha)$, there is $1+alpha^3+alpha^4$ and $1+alpha+alpha^2+alpha^3+alpha^4$.
    $endgroup$
    – TonyK
    Dec 13 '18 at 12:41














  • 1




    $begingroup$
    You basically compute the inverse of 1111 in the field and then multiply 0111 with the inverse of 1111.
    $endgroup$
    – Wuestenfux
    Dec 13 '18 at 11:35










  • $begingroup$
    So I need to calculate $1111^-1$, the problem is that this process in $GF(16)$ is not clear
    $endgroup$
    – Alessar
    Dec 13 '18 at 11:51






  • 1




    $begingroup$
    The extended euclidean algorithm gives a very clear way of doing it.
    $endgroup$
    – ancientmathematician
    Dec 13 '18 at 11:52






  • 1




    $begingroup$
    There is only one $GF(16)$, in the sense that all fields of 16 elements are isomorphic to each other. But there are many ways to represent an element of $GF(16)$, which depend on the specific irreducible polynomial used. So your question as it stands is badly specified. It is true that you mention $Pi(alpha)$, but only as an afterthought; whereas without it, we can't know what you mean by $0111$ and $1111$.
    $endgroup$
    – TonyK
    Dec 13 '18 at 12:34






  • 2




    $begingroup$
    When I said "many", I should in fact have said "three". In addition to your $Pi(alpha)$, there is $1+alpha^3+alpha^4$ and $1+alpha+alpha^2+alpha^3+alpha^4$.
    $endgroup$
    – TonyK
    Dec 13 '18 at 12:41








1




1




$begingroup$
You basically compute the inverse of 1111 in the field and then multiply 0111 with the inverse of 1111.
$endgroup$
– Wuestenfux
Dec 13 '18 at 11:35




$begingroup$
You basically compute the inverse of 1111 in the field and then multiply 0111 with the inverse of 1111.
$endgroup$
– Wuestenfux
Dec 13 '18 at 11:35












$begingroup$
So I need to calculate $1111^-1$, the problem is that this process in $GF(16)$ is not clear
$endgroup$
– Alessar
Dec 13 '18 at 11:51




$begingroup$
So I need to calculate $1111^-1$, the problem is that this process in $GF(16)$ is not clear
$endgroup$
– Alessar
Dec 13 '18 at 11:51




1




1




$begingroup$
The extended euclidean algorithm gives a very clear way of doing it.
$endgroup$
– ancientmathematician
Dec 13 '18 at 11:52




$begingroup$
The extended euclidean algorithm gives a very clear way of doing it.
$endgroup$
– ancientmathematician
Dec 13 '18 at 11:52




1




1




$begingroup$
There is only one $GF(16)$, in the sense that all fields of 16 elements are isomorphic to each other. But there are many ways to represent an element of $GF(16)$, which depend on the specific irreducible polynomial used. So your question as it stands is badly specified. It is true that you mention $Pi(alpha)$, but only as an afterthought; whereas without it, we can't know what you mean by $0111$ and $1111$.
$endgroup$
– TonyK
Dec 13 '18 at 12:34




$begingroup$
There is only one $GF(16)$, in the sense that all fields of 16 elements are isomorphic to each other. But there are many ways to represent an element of $GF(16)$, which depend on the specific irreducible polynomial used. So your question as it stands is badly specified. It is true that you mention $Pi(alpha)$, but only as an afterthought; whereas without it, we can't know what you mean by $0111$ and $1111$.
$endgroup$
– TonyK
Dec 13 '18 at 12:34




2




2




$begingroup$
When I said "many", I should in fact have said "three". In addition to your $Pi(alpha)$, there is $1+alpha^3+alpha^4$ and $1+alpha+alpha^2+alpha^3+alpha^4$.
$endgroup$
– TonyK
Dec 13 '18 at 12:41




$begingroup$
When I said "many", I should in fact have said "three". In addition to your $Pi(alpha)$, there is $1+alpha^3+alpha^4$ and $1+alpha+alpha^2+alpha^3+alpha^4$.
$endgroup$
– TonyK
Dec 13 '18 at 12:41










2 Answers
2






active

oldest

votes


















1












$begingroup$

As it happens, the discrete logarithm table for $GF(16)$ that I prepared for referrals like this, uses the same minimal polynomial of the primitive element. The only difference is that in the linked thread I denote by $gamma$ the element that you refer to as $alpha$.



Anyway, consulting that table, we see that
$$1111=1+alpha+alpha^2+alpha^3=alpha^{12},$$
and
$$
0111=alpha+alpha^2+alpha^3=alpha^{11}.
$$
Therefore
$$
begin{aligned}
frac{0111}{1111}&=frac{alpha^{11}}{alpha^{12}}=frac{alpha^{11}}{alpha^{12}}cdotfrac{alpha^3}{alpha^3}\
&=frac{alpha^{14}}{alpha^{15}}=frac{alpha^{14}}1\
&=alpha^{14}=alpha^3+1=1001.
end{aligned}
$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $GF(16)$ has characteristic 2. That is, each coefficient of $alpha$ is either $0$ or $1$. And $-1=1$.



    However, the polynomial division does not just result in $-1$ or $1$. Instead we have:
    $$frac{alpha^3+alpha^2+alpha}{alpha^3+alpha^2+alpha+1}
    = 1 + frac{1}{alpha^3+alpha^2+alpha+1}
    $$



    As an alternative to the answers in the comments, we can write each element (except $0$) as a power of $alpha$. After all, $GF(16)$ has a cyclic multiplicative group and $alpha^{15}=1$.



    Over the polynomial $X^4+X+1$ we have $0111 = alpha^{11}$ and $1111=alpha^{12}$. Therefore:
    $$0111/1111=alpha^{11}/alpha^{12}=alpha^{11+15-12}=alpha^{14}=1001$$



    This is consistent with your result:
    $$1 + frac{1}{alpha^3+alpha^2+alpha+1}=1+frac{1}{alpha^{12}}
    =1+frac{alpha^{15}}{alpha^{12}}=1+alpha^3
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Good work without a log table! If prompted for an ad hoc way of finding the discrete log of $1+alpha+alpha^2+alpha^3$ I would do the following $$1+alpha+alpha^2+alpha^3=(1+alpha)^3=(alpha^4)^3=alpha^{12}.$$
      $endgroup$
      – Jyrki Lahtonen
      Dec 13 '18 at 17:25













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037895%2fcalculations-on-gf16-find-0111-1111%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    As it happens, the discrete logarithm table for $GF(16)$ that I prepared for referrals like this, uses the same minimal polynomial of the primitive element. The only difference is that in the linked thread I denote by $gamma$ the element that you refer to as $alpha$.



    Anyway, consulting that table, we see that
    $$1111=1+alpha+alpha^2+alpha^3=alpha^{12},$$
    and
    $$
    0111=alpha+alpha^2+alpha^3=alpha^{11}.
    $$
    Therefore
    $$
    begin{aligned}
    frac{0111}{1111}&=frac{alpha^{11}}{alpha^{12}}=frac{alpha^{11}}{alpha^{12}}cdotfrac{alpha^3}{alpha^3}\
    &=frac{alpha^{14}}{alpha^{15}}=frac{alpha^{14}}1\
    &=alpha^{14}=alpha^3+1=1001.
    end{aligned}
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As it happens, the discrete logarithm table for $GF(16)$ that I prepared for referrals like this, uses the same minimal polynomial of the primitive element. The only difference is that in the linked thread I denote by $gamma$ the element that you refer to as $alpha$.



      Anyway, consulting that table, we see that
      $$1111=1+alpha+alpha^2+alpha^3=alpha^{12},$$
      and
      $$
      0111=alpha+alpha^2+alpha^3=alpha^{11}.
      $$
      Therefore
      $$
      begin{aligned}
      frac{0111}{1111}&=frac{alpha^{11}}{alpha^{12}}=frac{alpha^{11}}{alpha^{12}}cdotfrac{alpha^3}{alpha^3}\
      &=frac{alpha^{14}}{alpha^{15}}=frac{alpha^{14}}1\
      &=alpha^{14}=alpha^3+1=1001.
      end{aligned}
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As it happens, the discrete logarithm table for $GF(16)$ that I prepared for referrals like this, uses the same minimal polynomial of the primitive element. The only difference is that in the linked thread I denote by $gamma$ the element that you refer to as $alpha$.



        Anyway, consulting that table, we see that
        $$1111=1+alpha+alpha^2+alpha^3=alpha^{12},$$
        and
        $$
        0111=alpha+alpha^2+alpha^3=alpha^{11}.
        $$
        Therefore
        $$
        begin{aligned}
        frac{0111}{1111}&=frac{alpha^{11}}{alpha^{12}}=frac{alpha^{11}}{alpha^{12}}cdotfrac{alpha^3}{alpha^3}\
        &=frac{alpha^{14}}{alpha^{15}}=frac{alpha^{14}}1\
        &=alpha^{14}=alpha^3+1=1001.
        end{aligned}
        $$






        share|cite|improve this answer









        $endgroup$



        As it happens, the discrete logarithm table for $GF(16)$ that I prepared for referrals like this, uses the same minimal polynomial of the primitive element. The only difference is that in the linked thread I denote by $gamma$ the element that you refer to as $alpha$.



        Anyway, consulting that table, we see that
        $$1111=1+alpha+alpha^2+alpha^3=alpha^{12},$$
        and
        $$
        0111=alpha+alpha^2+alpha^3=alpha^{11}.
        $$
        Therefore
        $$
        begin{aligned}
        frac{0111}{1111}&=frac{alpha^{11}}{alpha^{12}}=frac{alpha^{11}}{alpha^{12}}cdotfrac{alpha^3}{alpha^3}\
        &=frac{alpha^{14}}{alpha^{15}}=frac{alpha^{14}}1\
        &=alpha^{14}=alpha^3+1=1001.
        end{aligned}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 12:23









        Jyrki LahtonenJyrki Lahtonen

        109k13169372




        109k13169372























            2












            $begingroup$

            $GF(16)$ has characteristic 2. That is, each coefficient of $alpha$ is either $0$ or $1$. And $-1=1$.



            However, the polynomial division does not just result in $-1$ or $1$. Instead we have:
            $$frac{alpha^3+alpha^2+alpha}{alpha^3+alpha^2+alpha+1}
            = 1 + frac{1}{alpha^3+alpha^2+alpha+1}
            $$



            As an alternative to the answers in the comments, we can write each element (except $0$) as a power of $alpha$. After all, $GF(16)$ has a cyclic multiplicative group and $alpha^{15}=1$.



            Over the polynomial $X^4+X+1$ we have $0111 = alpha^{11}$ and $1111=alpha^{12}$. Therefore:
            $$0111/1111=alpha^{11}/alpha^{12}=alpha^{11+15-12}=alpha^{14}=1001$$



            This is consistent with your result:
            $$1 + frac{1}{alpha^3+alpha^2+alpha+1}=1+frac{1}{alpha^{12}}
            =1+frac{alpha^{15}}{alpha^{12}}=1+alpha^3
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Good work without a log table! If prompted for an ad hoc way of finding the discrete log of $1+alpha+alpha^2+alpha^3$ I would do the following $$1+alpha+alpha^2+alpha^3=(1+alpha)^3=(alpha^4)^3=alpha^{12}.$$
              $endgroup$
              – Jyrki Lahtonen
              Dec 13 '18 at 17:25


















            2












            $begingroup$

            $GF(16)$ has characteristic 2. That is, each coefficient of $alpha$ is either $0$ or $1$. And $-1=1$.



            However, the polynomial division does not just result in $-1$ or $1$. Instead we have:
            $$frac{alpha^3+alpha^2+alpha}{alpha^3+alpha^2+alpha+1}
            = 1 + frac{1}{alpha^3+alpha^2+alpha+1}
            $$



            As an alternative to the answers in the comments, we can write each element (except $0$) as a power of $alpha$. After all, $GF(16)$ has a cyclic multiplicative group and $alpha^{15}=1$.



            Over the polynomial $X^4+X+1$ we have $0111 = alpha^{11}$ and $1111=alpha^{12}$. Therefore:
            $$0111/1111=alpha^{11}/alpha^{12}=alpha^{11+15-12}=alpha^{14}=1001$$



            This is consistent with your result:
            $$1 + frac{1}{alpha^3+alpha^2+alpha+1}=1+frac{1}{alpha^{12}}
            =1+frac{alpha^{15}}{alpha^{12}}=1+alpha^3
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Good work without a log table! If prompted for an ad hoc way of finding the discrete log of $1+alpha+alpha^2+alpha^3$ I would do the following $$1+alpha+alpha^2+alpha^3=(1+alpha)^3=(alpha^4)^3=alpha^{12}.$$
              $endgroup$
              – Jyrki Lahtonen
              Dec 13 '18 at 17:25
















            2












            2








            2





            $begingroup$

            $GF(16)$ has characteristic 2. That is, each coefficient of $alpha$ is either $0$ or $1$. And $-1=1$.



            However, the polynomial division does not just result in $-1$ or $1$. Instead we have:
            $$frac{alpha^3+alpha^2+alpha}{alpha^3+alpha^2+alpha+1}
            = 1 + frac{1}{alpha^3+alpha^2+alpha+1}
            $$



            As an alternative to the answers in the comments, we can write each element (except $0$) as a power of $alpha$. After all, $GF(16)$ has a cyclic multiplicative group and $alpha^{15}=1$.



            Over the polynomial $X^4+X+1$ we have $0111 = alpha^{11}$ and $1111=alpha^{12}$. Therefore:
            $$0111/1111=alpha^{11}/alpha^{12}=alpha^{11+15-12}=alpha^{14}=1001$$



            This is consistent with your result:
            $$1 + frac{1}{alpha^3+alpha^2+alpha+1}=1+frac{1}{alpha^{12}}
            =1+frac{alpha^{15}}{alpha^{12}}=1+alpha^3
            $$






            share|cite|improve this answer









            $endgroup$



            $GF(16)$ has characteristic 2. That is, each coefficient of $alpha$ is either $0$ or $1$. And $-1=1$.



            However, the polynomial division does not just result in $-1$ or $1$. Instead we have:
            $$frac{alpha^3+alpha^2+alpha}{alpha^3+alpha^2+alpha+1}
            = 1 + frac{1}{alpha^3+alpha^2+alpha+1}
            $$



            As an alternative to the answers in the comments, we can write each element (except $0$) as a power of $alpha$. After all, $GF(16)$ has a cyclic multiplicative group and $alpha^{15}=1$.



            Over the polynomial $X^4+X+1$ we have $0111 = alpha^{11}$ and $1111=alpha^{12}$. Therefore:
            $$0111/1111=alpha^{11}/alpha^{12}=alpha^{11+15-12}=alpha^{14}=1001$$



            This is consistent with your result:
            $$1 + frac{1}{alpha^3+alpha^2+alpha+1}=1+frac{1}{alpha^{12}}
            =1+frac{alpha^{15}}{alpha^{12}}=1+alpha^3
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 13 '18 at 12:25









            I like SerenaI like Serena

            4,1771722




            4,1771722












            • $begingroup$
              Good work without a log table! If prompted for an ad hoc way of finding the discrete log of $1+alpha+alpha^2+alpha^3$ I would do the following $$1+alpha+alpha^2+alpha^3=(1+alpha)^3=(alpha^4)^3=alpha^{12}.$$
              $endgroup$
              – Jyrki Lahtonen
              Dec 13 '18 at 17:25




















            • $begingroup$
              Good work without a log table! If prompted for an ad hoc way of finding the discrete log of $1+alpha+alpha^2+alpha^3$ I would do the following $$1+alpha+alpha^2+alpha^3=(1+alpha)^3=(alpha^4)^3=alpha^{12}.$$
              $endgroup$
              – Jyrki Lahtonen
              Dec 13 '18 at 17:25


















            $begingroup$
            Good work without a log table! If prompted for an ad hoc way of finding the discrete log of $1+alpha+alpha^2+alpha^3$ I would do the following $$1+alpha+alpha^2+alpha^3=(1+alpha)^3=(alpha^4)^3=alpha^{12}.$$
            $endgroup$
            – Jyrki Lahtonen
            Dec 13 '18 at 17:25






            $begingroup$
            Good work without a log table! If prompted for an ad hoc way of finding the discrete log of $1+alpha+alpha^2+alpha^3$ I would do the following $$1+alpha+alpha^2+alpha^3=(1+alpha)^3=(alpha^4)^3=alpha^{12}.$$
            $endgroup$
            – Jyrki Lahtonen
            Dec 13 '18 at 17:25




















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037895%2fcalculations-on-gf16-find-0111-1111%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more