Arriving at odd possible solutions for functions












0












$begingroup$


Say $f(x)=dfrac{3x+1}{2}$



I want to find out for which values of $x$ is the value of $f(x)$ an odd number.
So I reframe $f(x)$ to



$dfrac{3x+1}{2}=2k_1+1$



on simplifying further...



$dfrac{3x-1}{4}=k_1$ or $dfrac{4k_1+1}{3}=x$



For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.



Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.



The answer is $x=7,15,23,31...$
How do I frame the equation?



I tried putting



$dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and



$f(f(x))=f^2(x)= dfrac{3(frac{3x+1}{2})+1}{2}=2k_2+1$



and both of them lead to incorrect answers.






number-theory functions modular-arithmetic arithmetic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
    $endgroup$
    – lulu
    Dec 8 '18 at 10:52
















0












$begingroup$


Say $f(x)=dfrac{3x+1}{2}$



I want to find out for which values of $x$ is the value of $f(x)$ an odd number.
So I reframe $f(x)$ to



$dfrac{3x+1}{2}=2k_1+1$



on simplifying further...



$dfrac{3x-1}{4}=k_1$ or $dfrac{4k_1+1}{3}=x$



For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.



Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.



The answer is $x=7,15,23,31...$
How do I frame the equation?



I tried putting



$dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and



$f(f(x))=f^2(x)= dfrac{3(frac{3x+1}{2})+1}{2}=2k_2+1$



and both of them lead to incorrect answers.






number-theory functions modular-arithmetic arithmetic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
    $endgroup$
    – lulu
    Dec 8 '18 at 10:52














0












0








0





$begingroup$


Say $f(x)=dfrac{3x+1}{2}$



I want to find out for which values of $x$ is the value of $f(x)$ an odd number.
So I reframe $f(x)$ to



$dfrac{3x+1}{2}=2k_1+1$



on simplifying further...



$dfrac{3x-1}{4}=k_1$ or $dfrac{4k_1+1}{3}=x$



For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.



Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.



The answer is $x=7,15,23,31...$
How do I frame the equation?



I tried putting



$dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and



$f(f(x))=f^2(x)= dfrac{3(frac{3x+1}{2})+1}{2}=2k_2+1$



and both of them lead to incorrect answers.






number-theory functions modular-arithmetic arithmetic







share|cite|improve this question









$endgroup$




Say $f(x)=dfrac{3x+1}{2}$



I want to find out for which values of $x$ is the value of $f(x)$ an odd number.
So I reframe $f(x)$ to



$dfrac{3x+1}{2}=2k_1+1$



on simplifying further...



$dfrac{3x-1}{4}=k_1$ or $dfrac{4k_1+1}{3}=x$



For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.



Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.



The answer is $x=7,15,23,31...$
How do I frame the equation?



I tried putting



$dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and



$f(f(x))=f^2(x)= dfrac{3(frac{3x+1}{2})+1}{2}=2k_2+1$



and both of them lead to incorrect answers.






number-theory functions modular-arithmetic arithmetic




number-theory functions modular-arithmetic arithmetic






share|cite|improve this question













share|cite|improve this question











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asked Dec 8 '18 at 10:44









Rahul ShahRahul Shah

2202310




2202310








  • 1




    $begingroup$
    $fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
    $endgroup$
    – lulu
    Dec 8 '18 at 10:52














  • 1




    $begingroup$
    $fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
    $endgroup$
    – lulu
    Dec 8 '18 at 10:52








1




1




$begingroup$
$fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
$endgroup$
– lulu
Dec 8 '18 at 10:52




$begingroup$
$fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
$endgroup$
– lulu
Dec 8 '18 at 10:52










1 Answer
1






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$begingroup$

Continuing what you have written,



$f^2(x)=frac{9x+5}{4}=2k+1$



$k=frac{9x+1}{8}$, where k is any positive integer.



Note that $x=7$ satisfies this.



Further, let $x=7+y$ also gives integral $k$.



Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.



It gives $x=7,15,23,31...$






share|cite|improve this answer









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    $begingroup$

    Continuing what you have written,



    $f^2(x)=frac{9x+5}{4}=2k+1$



    $k=frac{9x+1}{8}$, where k is any positive integer.



    Note that $x=7$ satisfies this.



    Further, let $x=7+y$ also gives integral $k$.



    Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.



    It gives $x=7,15,23,31...$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Continuing what you have written,



      $f^2(x)=frac{9x+5}{4}=2k+1$



      $k=frac{9x+1}{8}$, where k is any positive integer.



      Note that $x=7$ satisfies this.



      Further, let $x=7+y$ also gives integral $k$.



      Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.



      It gives $x=7,15,23,31...$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Continuing what you have written,



        $f^2(x)=frac{9x+5}{4}=2k+1$



        $k=frac{9x+1}{8}$, where k is any positive integer.



        Note that $x=7$ satisfies this.



        Further, let $x=7+y$ also gives integral $k$.



        Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.



        It gives $x=7,15,23,31...$






        share|cite|improve this answer









        $endgroup$



        Continuing what you have written,



        $f^2(x)=frac{9x+5}{4}=2k+1$



        $k=frac{9x+1}{8}$, where k is any positive integer.



        Note that $x=7$ satisfies this.



        Further, let $x=7+y$ also gives integral $k$.



        Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.



        It gives $x=7,15,23,31...$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 10:56









        Ankit KumarAnkit Kumar

        1,494221




        1,494221






























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