Find the pairs $(m,k)$ solving the diophantine equation $m^2=7k+9$












0












$begingroup$


Solve



$$m^2=7k+9$$



over the integers



First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$



So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....






number-theory diophantine-equations integers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
    $endgroup$
    – Matti P.
    Jan 11 at 12:57








  • 2




    $begingroup$
    Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
    $endgroup$
    – Barry Cipra
    Jan 11 at 14:57










  • $begingroup$
    If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
    $endgroup$
    – rtybase
    Jan 11 at 17:49
















0












$begingroup$


Solve



$$m^2=7k+9$$



over the integers



First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$



So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....






number-theory diophantine-equations integers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
    $endgroup$
    – Matti P.
    Jan 11 at 12:57








  • 2




    $begingroup$
    Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
    $endgroup$
    – Barry Cipra
    Jan 11 at 14:57










  • $begingroup$
    If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
    $endgroup$
    – rtybase
    Jan 11 at 17:49














0












0








0





$begingroup$


Solve



$$m^2=7k+9$$



over the integers



First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$



So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....






number-theory diophantine-equations integers







share|cite|improve this question











$endgroup$




Solve



$$m^2=7k+9$$



over the integers



First i rearrange got $m^2-9=7k$
And $(m^2-9)/7=k$
So first $m^2-9$ must be divisible by $7$



So suppose $m=7n , 7n+1 ,...,7n+5$ But it doesn't work
....






number-theory diophantine-equations integers




number-theory diophantine-equations integers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 14:48









Peter

48.5k1139135




48.5k1139135










asked Jan 11 at 12:53









HeartHeart

29518




29518








  • 1




    $begingroup$
    So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
    $endgroup$
    – Matti P.
    Jan 11 at 12:57








  • 2




    $begingroup$
    Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
    $endgroup$
    – Barry Cipra
    Jan 11 at 14:57










  • $begingroup$
    If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
    $endgroup$
    – rtybase
    Jan 11 at 17:49














  • 1




    $begingroup$
    So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
    $endgroup$
    – Matti P.
    Jan 11 at 12:57








  • 2




    $begingroup$
    Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
    $endgroup$
    – Barry Cipra
    Jan 11 at 14:57










  • $begingroup$
    If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
    $endgroup$
    – rtybase
    Jan 11 at 17:49








1




1




$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57






$begingroup$
So what you're looking at is $$ m^2 equiv 2 {mod 7} $$ I would suggest plugging in some values for $m$ and seeing what happens. $m=1, 2, 3, 4, 5, 6, 7 ldots$. If you find a solution $m_0$,then $m_0 pm 7$ is also a solution. Therefore, you only need to look at the first $7$ integers.
$endgroup$
– Matti P.
Jan 11 at 12:57






2




2




$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57




$begingroup$
Hint: $m^2-9=(m-3)(m+3)$, so $7mid m^2-9iff7mid m-3lor7mid m+3$, since $7$ is prime.
$endgroup$
– Barry Cipra
Jan 11 at 14:57












$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49




$begingroup$
If $(m_0,k_0)$ is a solution, then $(m_0+7, k_0+2m_0+7)$ is also a solution $$m_0^2=7k_0+9 iff m_0^2+14m_0+49=7k_0+9+14m_0+49=7(k_0+2m_0+7)+9 iff\ (m_0+7)^2=7(k_0+2m_0+7)+9$$ and start with $(3,0)$.
$endgroup$
– rtybase
Jan 11 at 17:49










3 Answers
3






active

oldest

votes


















2












$begingroup$

$m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



$min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



For instance, if $m=81=4+77$, then $k=936$, and



$$81^2=7 cdot 936 + 9$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Above equation $m^2=7k+9$ has solution's,



    $m=7w-4$



    $k=(7w-1)(w-1)$, where, 'w' is any integer



    For $w=2$ we get, $(m.k)=(10,13)$



    $(10)^2=7(13)+9$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      What about (3,0)?
      $endgroup$
      – pendermath
      Jan 11 at 17:27










    • $begingroup$
      It is included $w=1$
      $endgroup$
      – Mike
      Jan 11 at 18:31










    • $begingroup$
      Sorry, I meant (4,1)
      $endgroup$
      – pendermath
      Jan 11 at 21:04










    • $begingroup$
      For w=0, you will get (m,k)=(4,1)
      $endgroup$
      – Sam
      Jan 11 at 22:10










    • $begingroup$
      If $w=0$, then $m=-4$, not 4, right?
      $endgroup$
      – pendermath
      Jan 12 at 22:37



















    0












    $begingroup$

    As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



    Therefore, you have two sets of solutions:



    Case 1: $7|m-3$ then
    $$m-3=7l \
    m=7l+3 \
    7k+9=m^2=49l^2+42l+9 \
    k=7l^2+6l\
    (k,m)= ( 7l^2+6l, 7l+3)$$



    Case 2: $7|m+3$ then
    $$m+3=7l \
    m=7l-3 \
    7k+9=m^2=49l^2-42l+9 \
    k=7l^2+6l\
    (k,m)= ( 7l^2-6l, 7l-3)$$






    share|cite|improve this answer









    $endgroup$













      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%2f3069798%2ffind-the-pairs-m-k-solving-the-diophantine-equation-m2-7k9%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      $m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



      The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



      $min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



      For instance, if $m=81=4+77$, then $k=936$, and



      $$81^2=7 cdot 936 + 9$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



        The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



        $min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



        For instance, if $m=81=4+77$, then $k=936$, and



        $$81^2=7 cdot 936 + 9$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



          The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



          $min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



          For instance, if $m=81=4+77$, then $k=936$, and



          $$81^2=7 cdot 936 + 9$$






          share|cite|improve this answer









          $endgroup$



          $m^2=7k+9 Rightarrow m^2=2 ;(mod 7)$.



          The only numbers from 1 to 7 satisfying the equation are 3 and 4 ($3^2=9=2; 4^2=16=2$. Then, the solution is of the form:



          $min{3+7n, 4+7n, n in mathbb{N}}$; and $k=frac{m^2-9}{7}$



          For instance, if $m=81=4+77$, then $k=936$, and



          $$81^2=7 cdot 936 + 9$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 17:22









          pendermathpendermath

          58612




          58612























              0












              $begingroup$

              Above equation $m^2=7k+9$ has solution's,



              $m=7w-4$



              $k=(7w-1)(w-1)$, where, 'w' is any integer



              For $w=2$ we get, $(m.k)=(10,13)$



              $(10)^2=7(13)+9$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                What about (3,0)?
                $endgroup$
                – pendermath
                Jan 11 at 17:27










              • $begingroup$
                It is included $w=1$
                $endgroup$
                – Mike
                Jan 11 at 18:31










              • $begingroup$
                Sorry, I meant (4,1)
                $endgroup$
                – pendermath
                Jan 11 at 21:04










              • $begingroup$
                For w=0, you will get (m,k)=(4,1)
                $endgroup$
                – Sam
                Jan 11 at 22:10










              • $begingroup$
                If $w=0$, then $m=-4$, not 4, right?
                $endgroup$
                – pendermath
                Jan 12 at 22:37
















              0












              $begingroup$

              Above equation $m^2=7k+9$ has solution's,



              $m=7w-4$



              $k=(7w-1)(w-1)$, where, 'w' is any integer



              For $w=2$ we get, $(m.k)=(10,13)$



              $(10)^2=7(13)+9$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                What about (3,0)?
                $endgroup$
                – pendermath
                Jan 11 at 17:27










              • $begingroup$
                It is included $w=1$
                $endgroup$
                – Mike
                Jan 11 at 18:31










              • $begingroup$
                Sorry, I meant (4,1)
                $endgroup$
                – pendermath
                Jan 11 at 21:04










              • $begingroup$
                For w=0, you will get (m,k)=(4,1)
                $endgroup$
                – Sam
                Jan 11 at 22:10










              • $begingroup$
                If $w=0$, then $m=-4$, not 4, right?
                $endgroup$
                – pendermath
                Jan 12 at 22:37














              0












              0








              0





              $begingroup$

              Above equation $m^2=7k+9$ has solution's,



              $m=7w-4$



              $k=(7w-1)(w-1)$, where, 'w' is any integer



              For $w=2$ we get, $(m.k)=(10,13)$



              $(10)^2=7(13)+9$






              share|cite|improve this answer











              $endgroup$



              Above equation $m^2=7k+9$ has solution's,



              $m=7w-4$



              $k=(7w-1)(w-1)$, where, 'w' is any integer



              For $w=2$ we get, $(m.k)=(10,13)$



              $(10)^2=7(13)+9$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 12 at 23:18









              Namaste

              1




              1










              answered Jan 11 at 17:26









              SamSam

              11




              11












              • $begingroup$
                What about (3,0)?
                $endgroup$
                – pendermath
                Jan 11 at 17:27










              • $begingroup$
                It is included $w=1$
                $endgroup$
                – Mike
                Jan 11 at 18:31










              • $begingroup$
                Sorry, I meant (4,1)
                $endgroup$
                – pendermath
                Jan 11 at 21:04










              • $begingroup$
                For w=0, you will get (m,k)=(4,1)
                $endgroup$
                – Sam
                Jan 11 at 22:10










              • $begingroup$
                If $w=0$, then $m=-4$, not 4, right?
                $endgroup$
                – pendermath
                Jan 12 at 22:37


















              • $begingroup$
                What about (3,0)?
                $endgroup$
                – pendermath
                Jan 11 at 17:27










              • $begingroup$
                It is included $w=1$
                $endgroup$
                – Mike
                Jan 11 at 18:31










              • $begingroup$
                Sorry, I meant (4,1)
                $endgroup$
                – pendermath
                Jan 11 at 21:04










              • $begingroup$
                For w=0, you will get (m,k)=(4,1)
                $endgroup$
                – Sam
                Jan 11 at 22:10










              • $begingroup$
                If $w=0$, then $m=-4$, not 4, right?
                $endgroup$
                – pendermath
                Jan 12 at 22:37
















              $begingroup$
              What about (3,0)?
              $endgroup$
              – pendermath
              Jan 11 at 17:27




              $begingroup$
              What about (3,0)?
              $endgroup$
              – pendermath
              Jan 11 at 17:27












              $begingroup$
              It is included $w=1$
              $endgroup$
              – Mike
              Jan 11 at 18:31




              $begingroup$
              It is included $w=1$
              $endgroup$
              – Mike
              Jan 11 at 18:31












              $begingroup$
              Sorry, I meant (4,1)
              $endgroup$
              – pendermath
              Jan 11 at 21:04




              $begingroup$
              Sorry, I meant (4,1)
              $endgroup$
              – pendermath
              Jan 11 at 21:04












              $begingroup$
              For w=0, you will get (m,k)=(4,1)
              $endgroup$
              – Sam
              Jan 11 at 22:10




              $begingroup$
              For w=0, you will get (m,k)=(4,1)
              $endgroup$
              – Sam
              Jan 11 at 22:10












              $begingroup$
              If $w=0$, then $m=-4$, not 4, right?
              $endgroup$
              – pendermath
              Jan 12 at 22:37




              $begingroup$
              If $w=0$, then $m=-4$, not 4, right?
              $endgroup$
              – pendermath
              Jan 12 at 22:37











              0












              $begingroup$

              As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



              Therefore, you have two sets of solutions:



              Case 1: $7|m-3$ then
              $$m-3=7l \
              m=7l+3 \
              7k+9=m^2=49l^2+42l+9 \
              k=7l^2+6l\
              (k,m)= ( 7l^2+6l, 7l+3)$$



              Case 2: $7|m+3$ then
              $$m+3=7l \
              m=7l-3 \
              7k+9=m^2=49l^2-42l+9 \
              k=7l^2+6l\
              (k,m)= ( 7l^2-6l, 7l-3)$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



                Therefore, you have two sets of solutions:



                Case 1: $7|m-3$ then
                $$m-3=7l \
                m=7l+3 \
                7k+9=m^2=49l^2+42l+9 \
                k=7l^2+6l\
                (k,m)= ( 7l^2+6l, 7l+3)$$



                Case 2: $7|m+3$ then
                $$m+3=7l \
                m=7l-3 \
                7k+9=m^2=49l^2-42l+9 \
                k=7l^2+6l\
                (k,m)= ( 7l^2-6l, 7l-3)$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



                  Therefore, you have two sets of solutions:



                  Case 1: $7|m-3$ then
                  $$m-3=7l \
                  m=7l+3 \
                  7k+9=m^2=49l^2+42l+9 \
                  k=7l^2+6l\
                  (k,m)= ( 7l^2+6l, 7l+3)$$



                  Case 2: $7|m+3$ then
                  $$m+3=7l \
                  m=7l-3 \
                  7k+9=m^2=49l^2-42l+9 \
                  k=7l^2+6l\
                  (k,m)= ( 7l^2-6l, 7l-3)$$






                  share|cite|improve this answer









                  $endgroup$



                  As you said $7$ must divide $m^2-9=(m-3)(m+3)$. Since $7$ is prime, it must divide either $m-3$ or $m+3$.



                  Therefore, you have two sets of solutions:



                  Case 1: $7|m-3$ then
                  $$m-3=7l \
                  m=7l+3 \
                  7k+9=m^2=49l^2+42l+9 \
                  k=7l^2+6l\
                  (k,m)= ( 7l^2+6l, 7l+3)$$



                  Case 2: $7|m+3$ then
                  $$m+3=7l \
                  m=7l-3 \
                  7k+9=m^2=49l^2-42l+9 \
                  k=7l^2+6l\
                  (k,m)= ( 7l^2-6l, 7l-3)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 12 at 23:45









                  N. S.N. S.

                  104k7114209




                  104k7114209






























                      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%2f3069798%2ffind-the-pairs-m-k-solving-the-diophantine-equation-m2-7k9%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

                      Human spaceflight

                      Image
                      Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
                      Read more

                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                      Image
                      0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
                      Read more

                      File:DeusFollowingSea.jpg

                      Image
                      File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
                      Read more