proof on difference of two squares and odd integers












0












$begingroup$


How would you prove that every odd integer is a difference of two squares?



I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.



So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it.
Could someone help me out please, thanks.






proof-writing







share|cite|improve this question









$endgroup$












  • $begingroup$
    See this picture: en.wikipedia.org/wiki/Square_number#Properties.
    $endgroup$
    – Michael Hoppe
    Dec 4 '18 at 19:35
















0












$begingroup$


How would you prove that every odd integer is a difference of two squares?



I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.



So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it.
Could someone help me out please, thanks.






proof-writing







share|cite|improve this question









$endgroup$












  • $begingroup$
    See this picture: en.wikipedia.org/wiki/Square_number#Properties.
    $endgroup$
    – Michael Hoppe
    Dec 4 '18 at 19:35














0












0








0





$begingroup$


How would you prove that every odd integer is a difference of two squares?



I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.



So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it.
Could someone help me out please, thanks.






proof-writing







share|cite|improve this question









$endgroup$




How would you prove that every odd integer is a difference of two squares?



I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.



So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it.
Could someone help me out please, thanks.






proof-writing




proof-writing






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 19:27









Carlos BaccaCarlos Bacca

180116




180116












  • $begingroup$
    See this picture: en.wikipedia.org/wiki/Square_number#Properties.
    $endgroup$
    – Michael Hoppe
    Dec 4 '18 at 19:35


















  • $begingroup$
    See this picture: en.wikipedia.org/wiki/Square_number#Properties.
    $endgroup$
    – Michael Hoppe
    Dec 4 '18 at 19:35
















$begingroup$
See this picture: en.wikipedia.org/wiki/Square_number#Properties.
$endgroup$
– Michael Hoppe
Dec 4 '18 at 19:35




$begingroup$
See this picture: en.wikipedia.org/wiki/Square_number#Properties.
$endgroup$
– Michael Hoppe
Dec 4 '18 at 19:35










5 Answers
5






active

oldest

votes


















0












$begingroup$

Hint: As you mentioned, you have $k = 2l+1$ for some $l$.



From here, note the difference of squares:



$$a^2-b^2 = (a+b)(a-b)$$



Let $a$ and $b$ be two consecutive integers and try to simplify.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
    $endgroup$
    – Carlos Bacca
    Dec 4 '18 at 19:38










  • $begingroup$
    You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
    $endgroup$
    – KM101
    Dec 4 '18 at 19:39












  • $begingroup$
    is that working backwards?
    $endgroup$
    – Carlos Bacca
    Dec 4 '18 at 19:40










  • $begingroup$
    actually ignore me
    $endgroup$
    – Carlos Bacca
    Dec 4 '18 at 19:41










  • $begingroup$
    Nope, it's the continuation of the proof.
    $endgroup$
    – KM101
    Dec 4 '18 at 19:41



















0












$begingroup$

Hint:



Compute $(n+1)^2-n^2$.




You should get $2n+1$







share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    HINT



    $$(k+1)^2-k^2=2k+1$$
    $$k^2-(k+1)^2=-(2k+1)$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Hint: What is the difference of squares of two consecutive numbers?






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        $$3=4-1$$



        $$2n+1=(n+1)^2-n^2$$






        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%2f3026022%2fproof-on-difference-of-two-squares-and-odd-integers%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Hint: As you mentioned, you have $k = 2l+1$ for some $l$.



          From here, note the difference of squares:



          $$a^2-b^2 = (a+b)(a-b)$$



          Let $a$ and $b$ be two consecutive integers and try to simplify.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:38










          • $begingroup$
            You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
            $endgroup$
            – KM101
            Dec 4 '18 at 19:39












          • $begingroup$
            is that working backwards?
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:40










          • $begingroup$
            actually ignore me
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:41










          • $begingroup$
            Nope, it's the continuation of the proof.
            $endgroup$
            – KM101
            Dec 4 '18 at 19:41
















          0












          $begingroup$

          Hint: As you mentioned, you have $k = 2l+1$ for some $l$.



          From here, note the difference of squares:



          $$a^2-b^2 = (a+b)(a-b)$$



          Let $a$ and $b$ be two consecutive integers and try to simplify.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:38










          • $begingroup$
            You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
            $endgroup$
            – KM101
            Dec 4 '18 at 19:39












          • $begingroup$
            is that working backwards?
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:40










          • $begingroup$
            actually ignore me
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:41










          • $begingroup$
            Nope, it's the continuation of the proof.
            $endgroup$
            – KM101
            Dec 4 '18 at 19:41














          0












          0








          0





          $begingroup$

          Hint: As you mentioned, you have $k = 2l+1$ for some $l$.



          From here, note the difference of squares:



          $$a^2-b^2 = (a+b)(a-b)$$



          Let $a$ and $b$ be two consecutive integers and try to simplify.






          share|cite|improve this answer









          $endgroup$



          Hint: As you mentioned, you have $k = 2l+1$ for some $l$.



          From here, note the difference of squares:



          $$a^2-b^2 = (a+b)(a-b)$$



          Let $a$ and $b$ be two consecutive integers and try to simplify.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 19:32









          KM101KM101

          5,9131523




          5,9131523












          • $begingroup$
            Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:38










          • $begingroup$
            You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
            $endgroup$
            – KM101
            Dec 4 '18 at 19:39












          • $begingroup$
            is that working backwards?
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:40










          • $begingroup$
            actually ignore me
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:41










          • $begingroup$
            Nope, it's the continuation of the proof.
            $endgroup$
            – KM101
            Dec 4 '18 at 19:41


















          • $begingroup$
            Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:38










          • $begingroup$
            You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
            $endgroup$
            – KM101
            Dec 4 '18 at 19:39












          • $begingroup$
            is that working backwards?
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:40










          • $begingroup$
            actually ignore me
            $endgroup$
            – Carlos Bacca
            Dec 4 '18 at 19:41










          • $begingroup$
            Nope, it's the continuation of the proof.
            $endgroup$
            – KM101
            Dec 4 '18 at 19:41
















          $begingroup$
          Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
          $endgroup$
          – Carlos Bacca
          Dec 4 '18 at 19:38




          $begingroup$
          Thanks, would you have to work from 2l+1 and then add $+l^2-l^2$ then factorise
          $endgroup$
          – Carlos Bacca
          Dec 4 '18 at 19:38












          $begingroup$
          You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
          $endgroup$
          – KM101
          Dec 4 '18 at 19:39






          $begingroup$
          You need to simplify $(b+1)^2-b^2$. Try using the difference of squares and simplify the expression. It will lead straight to the answer you want.
          $endgroup$
          – KM101
          Dec 4 '18 at 19:39














          $begingroup$
          is that working backwards?
          $endgroup$
          – Carlos Bacca
          Dec 4 '18 at 19:40




          $begingroup$
          is that working backwards?
          $endgroup$
          – Carlos Bacca
          Dec 4 '18 at 19:40












          $begingroup$
          actually ignore me
          $endgroup$
          – Carlos Bacca
          Dec 4 '18 at 19:41




          $begingroup$
          actually ignore me
          $endgroup$
          – Carlos Bacca
          Dec 4 '18 at 19:41












          $begingroup$
          Nope, it's the continuation of the proof.
          $endgroup$
          – KM101
          Dec 4 '18 at 19:41




          $begingroup$
          Nope, it's the continuation of the proof.
          $endgroup$
          – KM101
          Dec 4 '18 at 19:41











          0












          $begingroup$

          Hint:



          Compute $(n+1)^2-n^2$.




          You should get $2n+1$







          share|cite|improve this answer









          $endgroup$


















            0












            $begingroup$

            Hint:



            Compute $(n+1)^2-n^2$.




            You should get $2n+1$







            share|cite|improve this answer









            $endgroup$
















              0












              0








              0





              $begingroup$

              Hint:



              Compute $(n+1)^2-n^2$.




              You should get $2n+1$







              share|cite|improve this answer









              $endgroup$



              Hint:



              Compute $(n+1)^2-n^2$.




              You should get $2n+1$








              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 4 '18 at 19:29









              MasonMason

              1,9551530




              1,9551530























                  0












                  $begingroup$

                  HINT



                  $$(k+1)^2-k^2=2k+1$$
                  $$k^2-(k+1)^2=-(2k+1)$$






                  share|cite|improve this answer









                  $endgroup$


















                    0












                    $begingroup$

                    HINT



                    $$(k+1)^2-k^2=2k+1$$
                    $$k^2-(k+1)^2=-(2k+1)$$






                    share|cite|improve this answer









                    $endgroup$
















                      0












                      0








                      0





                      $begingroup$

                      HINT



                      $$(k+1)^2-k^2=2k+1$$
                      $$k^2-(k+1)^2=-(2k+1)$$






                      share|cite|improve this answer









                      $endgroup$



                      HINT



                      $$(k+1)^2-k^2=2k+1$$
                      $$k^2-(k+1)^2=-(2k+1)$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 4 '18 at 19:29









                      Rhys HughesRhys Hughes

                      5,7761528




                      5,7761528























                          0












                          $begingroup$

                          Hint: What is the difference of squares of two consecutive numbers?






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Hint: What is the difference of squares of two consecutive numbers?






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Hint: What is the difference of squares of two consecutive numbers?






                              share|cite|improve this answer









                              $endgroup$



                              Hint: What is the difference of squares of two consecutive numbers?







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 4 '18 at 19:29









                              Vasily MitchVasily Mitch

                              1,69638




                              1,69638























                                  0












                                  $begingroup$

                                  $$3=4-1$$



                                  $$2n+1=(n+1)^2-n^2$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    $$3=4-1$$



                                    $$2n+1=(n+1)^2-n^2$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      $$3=4-1$$



                                      $$2n+1=(n+1)^2-n^2$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      $$3=4-1$$



                                      $$2n+1=(n+1)^2-n^2$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 4 '18 at 19:37









                                      hamam_Abdallahhamam_Abdallah

                                      38k21634




                                      38k21634






























                                          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%2f3026022%2fproof-on-difference-of-two-squares-and-odd-integers%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

                                          Probability when a professor distributes a quiz and homework assignment to a class of n students.

                                          Image
                                          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                                          Read more

                                          Aardman Animations

                                          Image
                                          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                                          Read more

                                          Are they similar matrix

                                          Image
                                          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                                          Read more