Proving that the sum and difference of two squares (not equal to zero) can't both be squares.












7














I have the following task:




Prove that the sum and the difference of two squares (not equal to zero) can't both be squares.




For the sum, I thought about Pythagorean triples: $x^2+y^2=z^2$ works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well?



I tried to write down this:



Assume that $a^2>b^2$. Then $a^2+b^2=c^2$ and $a^2-b^2=d^2$. Summing the left and the right equation, we get $2a^2=c^2+d^2$. I am just stuck proving why this can't occur. Any ideas? Thanks for your help!






elementary-number-theory







share|cite|improve this question





























    7














    I have the following task:




    Prove that the sum and the difference of two squares (not equal to zero) can't both be squares.




    For the sum, I thought about Pythagorean triples: $x^2+y^2=z^2$ works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well?



    I tried to write down this:



    Assume that $a^2>b^2$. Then $a^2+b^2=c^2$ and $a^2-b^2=d^2$. Summing the left and the right equation, we get $2a^2=c^2+d^2$. I am just stuck proving why this can't occur. Any ideas? Thanks for your help!






    elementary-number-theory







    share|cite|improve this question



























      7












      7








      7


      4





      I have the following task:




      Prove that the sum and the difference of two squares (not equal to zero) can't both be squares.




      For the sum, I thought about Pythagorean triples: $x^2+y^2=z^2$ works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well?



      I tried to write down this:



      Assume that $a^2>b^2$. Then $a^2+b^2=c^2$ and $a^2-b^2=d^2$. Summing the left and the right equation, we get $2a^2=c^2+d^2$. I am just stuck proving why this can't occur. Any ideas? Thanks for your help!






      elementary-number-theory







      share|cite|improve this question















      I have the following task:




      Prove that the sum and the difference of two squares (not equal to zero) can't both be squares.




      For the sum, I thought about Pythagorean triples: $x^2+y^2=z^2$ works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well?



      I tried to write down this:



      Assume that $a^2>b^2$. Then $a^2+b^2=c^2$ and $a^2-b^2=d^2$. Summing the left and the right equation, we get $2a^2=c^2+d^2$. I am just stuck proving why this can't occur. Any ideas? Thanks for your help!






      elementary-number-theory




      elementary-number-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 25 '15 at 17:34









      kate

      5281717




      5281717










      asked Apr 22 '15 at 11:13









      Atvin

      2,8451825




      2,8451825






















          2 Answers
          2






          active

          oldest

          votes


















          7














          If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



          It is known that this is impossible. See, for example, here.






          share|cite|improve this answer

















          • 1




            Very nice, thanks! :)
            – Atvin
            Apr 22 '15 at 11:19



















          3














          This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



          Fibonacci's Lost Theorem $ $ The area of an integral
          pythagorean triangle is not a perfect square.



          A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



          Naturally associated with every SAP is a "half square triangle",
          ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
          $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



          which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



          With these concepts in mind, the proof is very easy:
          If there exists a pythagorean triangle with square area then it
          may be primitivized and its area remains square. Let its primitive
          parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



          $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



          Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
          Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
          yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



          Remark $ $ This doubling construction is ancient - already in Euclid. It may be
          viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $






          share|cite|improve this answer





















            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%2f1246469%2fproving-that-the-sum-and-difference-of-two-squares-not-equal-to-zero-cant-bot%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









            7














            If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



            It is known that this is impossible. See, for example, here.






            share|cite|improve this answer

















            • 1




              Very nice, thanks! :)
              – Atvin
              Apr 22 '15 at 11:19
















            7














            If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



            It is known that this is impossible. See, for example, here.






            share|cite|improve this answer

















            • 1




              Very nice, thanks! :)
              – Atvin
              Apr 22 '15 at 11:19














            7












            7








            7






            If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



            It is known that this is impossible. See, for example, here.






            share|cite|improve this answer












            If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



            It is known that this is impossible. See, for example, here.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 22 '15 at 11:18









            mathlove

            91.7k881214




            91.7k881214








            • 1




              Very nice, thanks! :)
              – Atvin
              Apr 22 '15 at 11:19














            • 1




              Very nice, thanks! :)
              – Atvin
              Apr 22 '15 at 11:19








            1




            1




            Very nice, thanks! :)
            – Atvin
            Apr 22 '15 at 11:19




            Very nice, thanks! :)
            – Atvin
            Apr 22 '15 at 11:19











            3














            This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



            Fibonacci's Lost Theorem $ $ The area of an integral
            pythagorean triangle is not a perfect square.



            A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



            Naturally associated with every SAP is a "half square triangle",
            ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
            $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



            which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



            With these concepts in mind, the proof is very easy:
            If there exists a pythagorean triangle with square area then it
            may be primitivized and its area remains square. Let its primitive
            parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



            $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



            Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
            Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
            yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



            Remark $ $ This doubling construction is ancient - already in Euclid. It may be
            viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $






            share|cite|improve this answer


























              3














              This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



              Fibonacci's Lost Theorem $ $ The area of an integral
              pythagorean triangle is not a perfect square.



              A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



              Naturally associated with every SAP is a "half square triangle",
              ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
              $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



              which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



              With these concepts in mind, the proof is very easy:
              If there exists a pythagorean triangle with square area then it
              may be primitivized and its area remains square. Let its primitive
              parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



              $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



              Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
              Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
              yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



              Remark $ $ This doubling construction is ancient - already in Euclid. It may be
              viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $






              share|cite|improve this answer
























                3












                3








                3






                This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



                Fibonacci's Lost Theorem $ $ The area of an integral
                pythagorean triangle is not a perfect square.



                A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



                Naturally associated with every SAP is a "half square triangle",
                ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
                $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



                which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



                With these concepts in mind, the proof is very easy:
                If there exists a pythagorean triangle with square area then it
                may be primitivized and its area remains square. Let its primitive
                parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



                $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



                Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
                Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
                yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



                Remark $ $ This doubling construction is ancient - already in Euclid. It may be
                viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $






                share|cite|improve this answer












                This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



                Fibonacci's Lost Theorem $ $ The area of an integral
                pythagorean triangle is not a perfect square.



                A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



                Naturally associated with every SAP is a "half square triangle",
                ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
                $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



                which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



                With these concepts in mind, the proof is very easy:
                If there exists a pythagorean triangle with square area then it
                may be primitivized and its area remains square. Let its primitive
                parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



                $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



                Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
                Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
                yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



                Remark $ $ This doubling construction is ancient - already in Euclid. It may be
                viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered May 4 '15 at 16:38









                Bill Dubuque

                208k29190627




                208k29190627






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f1246469%2fproving-that-the-sum-and-difference-of-two-squares-not-equal-to-zero-cant-bot%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

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                    Read more

                    In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
                    Read more

                    How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

                    Image
                    1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
                    Read more