Is it possible for a square root function,f(x), to map to a finite number of integers for all x in domain of...












2












$begingroup$


Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$



Using python I checked x for $$ -100000000 leq x leq 100000000$$



and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?



Edit: $$x in mathbb{Z}$$






elementary-number-theory







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$endgroup$








  • 4




    $begingroup$
    To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
    $endgroup$
    – Théophile
    Apr 5 at 20:08






  • 1




    $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
    $endgroup$
    – Arturo Magidin
    Apr 5 at 20:11










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbb{Z}$$
    $endgroup$
    – Diehardwalnut
    Apr 5 at 21:28
















2












$begingroup$


Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$



Using python I checked x for $$ -100000000 leq x leq 100000000$$



and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?



Edit: $$x in mathbb{Z}$$






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
    $endgroup$
    – Théophile
    Apr 5 at 20:08






  • 1




    $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
    $endgroup$
    – Arturo Magidin
    Apr 5 at 20:11










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbb{Z}$$
    $endgroup$
    – Diehardwalnut
    Apr 5 at 21:28














2












2








2


1



$begingroup$


Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$



Using python I checked x for $$ -100000000 leq x leq 100000000$$



and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?



Edit: $$x in mathbb{Z}$$






elementary-number-theory







share|cite|improve this question











$endgroup$




Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$



Using python I checked x for $$ -100000000 leq x leq 100000000$$



and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?



Edit: $$x in mathbb{Z}$$






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 5 at 21:27







Diehardwalnut

















asked Apr 5 at 20:04









DiehardwalnutDiehardwalnut

262110




262110








  • 4




    $begingroup$
    To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
    $endgroup$
    – Théophile
    Apr 5 at 20:08






  • 1




    $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
    $endgroup$
    – Arturo Magidin
    Apr 5 at 20:11










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbb{Z}$$
    $endgroup$
    – Diehardwalnut
    Apr 5 at 21:28














  • 4




    $begingroup$
    To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
    $endgroup$
    – Théophile
    Apr 5 at 20:08






  • 1




    $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
    $endgroup$
    – Arturo Magidin
    Apr 5 at 20:11










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbb{Z}$$
    $endgroup$
    – Diehardwalnut
    Apr 5 at 21:28








4




4




$begingroup$
To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
$endgroup$
– Théophile
Apr 5 at 20:08




$begingroup$
To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
$endgroup$
– Théophile
Apr 5 at 20:08




1




1




$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
$endgroup$
– Arturo Magidin
Apr 5 at 20:11




$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
$endgroup$
– Arturo Magidin
Apr 5 at 20:11












$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
Apr 5 at 21:28




$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
Apr 5 at 21:28










3 Answers
3






active

oldest

votes


















4












$begingroup$

First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



Furthermore
begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
end{align*}



Can you end it now?




Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.







share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
    $$y^2 = x^2-x+1,$$
    so $y^2-1=x^2-x$. That is,
    $$(y+1)(y-1) = x(x-1).$$



    So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Hint:



      For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



      for $x<0$, $x^2<x^2-x+1<(x-1)^2$.






      share|cite|improve this answer









      $endgroup$














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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



        Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



        It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



        Furthermore
        begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
        x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
        end{align*}



        Can you end it now?




        Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.







        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



          Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



          It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



          Furthermore
          begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
          x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
          end{align*}



          Can you end it now?




          Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.







          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



            Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



            It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



            Furthermore
            begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
            x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
            end{align*}



            Can you end it now?




            Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.







            share|cite|improve this answer











            $endgroup$



            First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



            Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



            It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



            Furthermore
            begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
            x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
            end{align*}



            Can you end it now?




            Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 6 at 1:51









            Carmeister

            2,8692924




            2,8692924










            answered Apr 5 at 20:14









            Dr. MathvaDr. Mathva

            3,493630




            3,493630























                3












                $begingroup$

                Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
                $$y^2 = x^2-x+1,$$
                so $y^2-1=x^2-x$. That is,
                $$(y+1)(y-1) = x(x-1).$$



                So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
                  $$y^2 = x^2-x+1,$$
                  so $y^2-1=x^2-x$. That is,
                  $$(y+1)(y-1) = x(x-1).$$



                  So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
                    $$y^2 = x^2-x+1,$$
                    so $y^2-1=x^2-x$. That is,
                    $$(y+1)(y-1) = x(x-1).$$



                    So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?






                    share|cite|improve this answer









                    $endgroup$



                    Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
                    $$y^2 = x^2-x+1,$$
                    so $y^2-1=x^2-x$. That is,
                    $$(y+1)(y-1) = x(x-1).$$



                    So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Apr 5 at 20:14









                    ThéophileThéophile

                    20.4k13047




                    20.4k13047























                        2












                        $begingroup$

                        Hint:



                        For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



                        for $x<0$, $x^2<x^2-x+1<(x-1)^2$.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Hint:



                          For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



                          for $x<0$, $x^2<x^2-x+1<(x-1)^2$.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Hint:



                            For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



                            for $x<0$, $x^2<x^2-x+1<(x-1)^2$.






                            share|cite|improve this answer









                            $endgroup$



                            Hint:



                            For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



                            for $x<0$, $x^2<x^2-x+1<(x-1)^2$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Apr 5 at 21:49









                            J. W. TannerJ. W. Tanner

                            4,7721420




                            4,7721420






























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