Can 2018 be written as a sum of two squares? If can, what is the expression?












0












$begingroup$



Can $2018$ be written as a sum of two squares?



If it can be, what are the numbers?




I know the first answer is it can be because $2018=2times1009$ and $1009equiv 1 pmod{4}$. But I cannot find the numbers.






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If all you care about is the numbers, then wolframalpha can help.
    $endgroup$
    – JMoravitz
    Dec 7 '17 at 4:18










  • $begingroup$
    By hit and trial, you can verify that $2018 = 13^2 + cdots$
    $endgroup$
    – Math Lover
    Dec 7 '17 at 4:21










  • $begingroup$
    For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
    $endgroup$
    – Ross Millikan
    Dec 7 '17 at 4:21












  • $begingroup$
    See math.stackexchange.com/questions/1828694/…
    $endgroup$
    – lab bhattacharjee
    Dec 7 '17 at 4:41
















0












$begingroup$



Can $2018$ be written as a sum of two squares?



If it can be, what are the numbers?




I know the first answer is it can be because $2018=2times1009$ and $1009equiv 1 pmod{4}$. But I cannot find the numbers.






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If all you care about is the numbers, then wolframalpha can help.
    $endgroup$
    – JMoravitz
    Dec 7 '17 at 4:18










  • $begingroup$
    By hit and trial, you can verify that $2018 = 13^2 + cdots$
    $endgroup$
    – Math Lover
    Dec 7 '17 at 4:21










  • $begingroup$
    For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
    $endgroup$
    – Ross Millikan
    Dec 7 '17 at 4:21












  • $begingroup$
    See math.stackexchange.com/questions/1828694/…
    $endgroup$
    – lab bhattacharjee
    Dec 7 '17 at 4:41














0












0








0


2



$begingroup$



Can $2018$ be written as a sum of two squares?



If it can be, what are the numbers?




I know the first answer is it can be because $2018=2times1009$ and $1009equiv 1 pmod{4}$. But I cannot find the numbers.






elementary-number-theory







share|cite|improve this question











$endgroup$





Can $2018$ be written as a sum of two squares?



If it can be, what are the numbers?




I know the first answer is it can be because $2018=2times1009$ and $1009equiv 1 pmod{4}$. But I cannot find the numbers.






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 Dec 16 '17 at 17:39









Namaste

1




1










asked Dec 7 '17 at 4:15









Joe sae Joe sae

337




337








  • 1




    $begingroup$
    If all you care about is the numbers, then wolframalpha can help.
    $endgroup$
    – JMoravitz
    Dec 7 '17 at 4:18










  • $begingroup$
    By hit and trial, you can verify that $2018 = 13^2 + cdots$
    $endgroup$
    – Math Lover
    Dec 7 '17 at 4:21










  • $begingroup$
    For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
    $endgroup$
    – Ross Millikan
    Dec 7 '17 at 4:21












  • $begingroup$
    See math.stackexchange.com/questions/1828694/…
    $endgroup$
    – lab bhattacharjee
    Dec 7 '17 at 4:41














  • 1




    $begingroup$
    If all you care about is the numbers, then wolframalpha can help.
    $endgroup$
    – JMoravitz
    Dec 7 '17 at 4:18










  • $begingroup$
    By hit and trial, you can verify that $2018 = 13^2 + cdots$
    $endgroup$
    – Math Lover
    Dec 7 '17 at 4:21










  • $begingroup$
    For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
    $endgroup$
    – Ross Millikan
    Dec 7 '17 at 4:21












  • $begingroup$
    See math.stackexchange.com/questions/1828694/…
    $endgroup$
    – lab bhattacharjee
    Dec 7 '17 at 4:41








1




1




$begingroup$
If all you care about is the numbers, then wolframalpha can help.
$endgroup$
– JMoravitz
Dec 7 '17 at 4:18




$begingroup$
If all you care about is the numbers, then wolframalpha can help.
$endgroup$
– JMoravitz
Dec 7 '17 at 4:18












$begingroup$
By hit and trial, you can verify that $2018 = 13^2 + cdots$
$endgroup$
– Math Lover
Dec 7 '17 at 4:21




$begingroup$
By hit and trial, you can verify that $2018 = 13^2 + cdots$
$endgroup$
– Math Lover
Dec 7 '17 at 4:21












$begingroup$
For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
$endgroup$
– Ross Millikan
Dec 7 '17 at 4:21






$begingroup$
For a small number like this, just make a column in a spreadsheet with $1$ to $45$ then =sqrt(2018-left^2) will make it easy to find. Just search by eye for the entries without decimals. They leap out.
$endgroup$
– Ross Millikan
Dec 7 '17 at 4:21














$begingroup$
See math.stackexchange.com/questions/1828694/…
$endgroup$
– lab bhattacharjee
Dec 7 '17 at 4:41




$begingroup$
See math.stackexchange.com/questions/1828694/…
$endgroup$
– lab bhattacharjee
Dec 7 '17 at 4:41










2 Answers
2






active

oldest

votes


















5












$begingroup$

Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
Brahmagupta–Fibonacci identity
$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
tell us $2018 = (a+b)^2 + (a-b)^2$.



The task reduces to rewrite $1009$ as a sum of two squares. In addition to
seaching by brute force, one can use algorithms described in
answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$



A good reference of these sort of algorithms will be Henri Cohen's book
A course in Computation Algebraic Number Theory. Take a look at that if you need more details.






share|cite|improve this answer











$endgroup$





















    -1












    $begingroup$

    You are correct my friend,as per Brahmagupta's Fibonacci identity,it says that-$a^2c^2+a^2d^2+b^2c^2+b^2d^2$=$a^2c^2-2acbd+b^2d^2+a^2d^2+2adbc+b^2c^2$=$a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2$ just as you are writing 1009=784+225 and 2018=1849+169






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Now if you want to use only 43 and 13 to represent 1009 then you can write as follows:$$1009=[(frac{1}{4}){(43+13)^2+(43-13)^2}]$$= $$(frac{1}{4}){(56^2+30^2)$$ = $$(frac{1}{4})(3136+900)$$
      $endgroup$
      – user630998
      Jan 2 at 5:09














    Your Answer








    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%2f2555005%2fcan-2018-be-written-as-a-sum-of-two-squares-if-can-what-is-the-expression%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









    5












    $begingroup$

    Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
    Brahmagupta–Fibonacci identity
    $$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
    tell us $2018 = (a+b)^2 + (a-b)^2$.



    The task reduces to rewrite $1009$ as a sum of two squares. In addition to
    seaching by brute force, one can use algorithms described in
    answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$



    A good reference of these sort of algorithms will be Henri Cohen's book
    A course in Computation Algebraic Number Theory. Take a look at that if you need more details.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
      Brahmagupta–Fibonacci identity
      $$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
      tell us $2018 = (a+b)^2 + (a-b)^2$.



      The task reduces to rewrite $1009$ as a sum of two squares. In addition to
      seaching by brute force, one can use algorithms described in
      answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$



      A good reference of these sort of algorithms will be Henri Cohen's book
      A course in Computation Algebraic Number Theory. Take a look at that if you need more details.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
        Brahmagupta–Fibonacci identity
        $$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
        tell us $2018 = (a+b)^2 + (a-b)^2$.



        The task reduces to rewrite $1009$ as a sum of two squares. In addition to
        seaching by brute force, one can use algorithms described in
        answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$



        A good reference of these sort of algorithms will be Henri Cohen's book
        A course in Computation Algebraic Number Theory. Take a look at that if you need more details.






        share|cite|improve this answer











        $endgroup$



        Yes, $2018 = 2times 1009$ (as prime factorization). Since $1009$ is an odd prime of the form $4k+1$, it can be written as a sum of two squares $1009 = a^2+b^2$. Together with $2 = 1^2 + 1^2$,
        Brahmagupta–Fibonacci identity
        $$(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$
        tell us $2018 = (a+b)^2 + (a-b)^2$.



        The task reduces to rewrite $1009$ as a sum of two squares. In addition to
        seaching by brute force, one can use algorithms described in
        answers of this question to determine $a,b$ efficiently. At the end, one find $$1009 = 28^2+15^2 quadimpliesquad 2018 = 43^2 + 13^2$$



        A good reference of these sort of algorithms will be Henri Cohen's book
        A course in Computation Algebraic Number Theory. Take a look at that if you need more details.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 '18 at 19:49









        LoPablo

        33




        33










        answered Dec 7 '17 at 4:51









        achille huiachille hui

        96.7k5132261




        96.7k5132261























            -1












            $begingroup$

            You are correct my friend,as per Brahmagupta's Fibonacci identity,it says that-$a^2c^2+a^2d^2+b^2c^2+b^2d^2$=$a^2c^2-2acbd+b^2d^2+a^2d^2+2adbc+b^2c^2$=$a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2$ just as you are writing 1009=784+225 and 2018=1849+169






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now if you want to use only 43 and 13 to represent 1009 then you can write as follows:$$1009=[(frac{1}{4}){(43+13)^2+(43-13)^2}]$$= $$(frac{1}{4}){(56^2+30^2)$$ = $$(frac{1}{4})(3136+900)$$
              $endgroup$
              – user630998
              Jan 2 at 5:09


















            -1












            $begingroup$

            You are correct my friend,as per Brahmagupta's Fibonacci identity,it says that-$a^2c^2+a^2d^2+b^2c^2+b^2d^2$=$a^2c^2-2acbd+b^2d^2+a^2d^2+2adbc+b^2c^2$=$a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2$ just as you are writing 1009=784+225 and 2018=1849+169






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now if you want to use only 43 and 13 to represent 1009 then you can write as follows:$$1009=[(frac{1}{4}){(43+13)^2+(43-13)^2}]$$= $$(frac{1}{4}){(56^2+30^2)$$ = $$(frac{1}{4})(3136+900)$$
              $endgroup$
              – user630998
              Jan 2 at 5:09
















            -1












            -1








            -1





            $begingroup$

            You are correct my friend,as per Brahmagupta's Fibonacci identity,it says that-$a^2c^2+a^2d^2+b^2c^2+b^2d^2$=$a^2c^2-2acbd+b^2d^2+a^2d^2+2adbc+b^2c^2$=$a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2$ just as you are writing 1009=784+225 and 2018=1849+169






            share|cite|improve this answer









            $endgroup$



            You are correct my friend,as per Brahmagupta's Fibonacci identity,it says that-$a^2c^2+a^2d^2+b^2c^2+b^2d^2$=$a^2c^2-2acbd+b^2d^2+a^2d^2+2adbc+b^2c^2$=$a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2$ just as you are writing 1009=784+225 and 2018=1849+169







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 2 at 4:55







            user630998



















            • $begingroup$
              Now if you want to use only 43 and 13 to represent 1009 then you can write as follows:$$1009=[(frac{1}{4}){(43+13)^2+(43-13)^2}]$$= $$(frac{1}{4}){(56^2+30^2)$$ = $$(frac{1}{4})(3136+900)$$
              $endgroup$
              – user630998
              Jan 2 at 5:09




















            • $begingroup$
              Now if you want to use only 43 and 13 to represent 1009 then you can write as follows:$$1009=[(frac{1}{4}){(43+13)^2+(43-13)^2}]$$= $$(frac{1}{4}){(56^2+30^2)$$ = $$(frac{1}{4})(3136+900)$$
              $endgroup$
              – user630998
              Jan 2 at 5:09


















            $begingroup$
            Now if you want to use only 43 and 13 to represent 1009 then you can write as follows:$$1009=[(frac{1}{4}){(43+13)^2+(43-13)^2}]$$= $$(frac{1}{4}){(56^2+30^2)$$ = $$(frac{1}{4})(3136+900)$$
            $endgroup$
            – user630998
            Jan 2 at 5:09






            $begingroup$
            Now if you want to use only 43 and 13 to represent 1009 then you can write as follows:$$1009=[(frac{1}{4}){(43+13)^2+(43-13)^2}]$$= $$(frac{1}{4}){(56^2+30^2)$$ = $$(frac{1}{4})(3136+900)$$
            $endgroup$
            – user630998
            Jan 2 at 5:09




















            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%2f2555005%2fcan-2018-be-written-as-a-sum-of-two-squares-if-can-what-is-the-expression%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 to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more