Using these conditions to imply that $(1/π )arctan(1/3)$ is irrational












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Suppose $x_n equiv 3 pmod 5$ and $y_n equiv 1 pmod 5$ and that $(3 + i)^n = x_n + iy_n$, with n as an integer greater than or equal to $1$. How does this imply that $frac{1}{pi}arctan(frac{1}{3})$ is irrational?



I have found so far that $x_n+1 = 3x_n - y_n$
and $y_n+1 = x_n + 3y_n$,
where $x$ is the real part and $y$ is the imaginary part.










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  • $begingroup$
    Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
    $endgroup$
    – David
    Dec 4 '18 at 1:30
















-2












$begingroup$


Suppose $x_n equiv 3 pmod 5$ and $y_n equiv 1 pmod 5$ and that $(3 + i)^n = x_n + iy_n$, with n as an integer greater than or equal to $1$. How does this imply that $frac{1}{pi}arctan(frac{1}{3})$ is irrational?



I have found so far that $x_n+1 = 3x_n - y_n$
and $y_n+1 = x_n + 3y_n$,
where $x$ is the real part and $y$ is the imaginary part.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
    $endgroup$
    – David
    Dec 4 '18 at 1:30














-2












-2








-2


1



$begingroup$


Suppose $x_n equiv 3 pmod 5$ and $y_n equiv 1 pmod 5$ and that $(3 + i)^n = x_n + iy_n$, with n as an integer greater than or equal to $1$. How does this imply that $frac{1}{pi}arctan(frac{1}{3})$ is irrational?



I have found so far that $x_n+1 = 3x_n - y_n$
and $y_n+1 = x_n + 3y_n$,
where $x$ is the real part and $y$ is the imaginary part.










share|cite|improve this question











$endgroup$




Suppose $x_n equiv 3 pmod 5$ and $y_n equiv 1 pmod 5$ and that $(3 + i)^n = x_n + iy_n$, with n as an integer greater than or equal to $1$. How does this imply that $frac{1}{pi}arctan(frac{1}{3})$ is irrational?



I have found so far that $x_n+1 = 3x_n - y_n$
and $y_n+1 = x_n + 3y_n$,
where $x$ is the real part and $y$ is the imaginary part.







complex-analysis






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edited Dec 4 '18 at 10:06









Tianlalu

3,08621038




3,08621038










asked Dec 4 '18 at 1:21









ReyRey

12




12












  • $begingroup$
    Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
    $endgroup$
    – David
    Dec 4 '18 at 1:30


















  • $begingroup$
    Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
    $endgroup$
    – David
    Dec 4 '18 at 1:30
















$begingroup$
Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
$endgroup$
– David
Dec 4 '18 at 1:30




$begingroup$
Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
$endgroup$
– David
Dec 4 '18 at 1:30










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You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.



Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.






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    $begingroup$

    You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.



    Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.






    share|cite|improve this answer









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      0












      $begingroup$

      You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.



      Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.



        Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.






        share|cite|improve this answer









        $endgroup$



        You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.



        Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 1:43









        Oscar LanziOscar Lanzi

        12.3k12036




        12.3k12036






























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