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

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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














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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    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




















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