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

                                          How do I know what Microsoft account the skydrive app is syncing to?

                                          Image
                                          up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
                                          Read more

                                          When does type information flow backwards in C++?

                                          Image
                                          up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
                                          Read more

                                          Grease: Live!

                                          Image
                                          A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
                                          Read more