Is it possible for a square root function,f(x), to map to a finite number of integers for all x in domain of f?For integers $a$ and $b gt 0$, and $n^2$ a sum of two square integers, does this strategy find the largest integer $x | x^2 lt n^2(a^2 + b^2)$?Do roots of a polynomial with coefficients from a Collatz sequence all fall in a disk of radius 1.5?A fun little problemWhat is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer?on roots of an equationAn integer sequence with integer $k$ normsProof Verification: If $x$ is a nonnegative real number, then $big[sqrt[x]big] = big[sqrtxbig]$Digit after decimal point of radicalsProving that there does not exist an infinite descending sequence of naturals using minimal counterexamplenumber of different ways to represent a positive integer as a binomial coefficient

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


For integers $a$ and $b gt 0$, and $n^2$ a sum of two square integers, does this strategy find the largest integer $x | x^2 lt n^2(a^2 + b^2)$?Do roots of a polynomial with coefficients from a Collatz sequence all fall in a disk of radius 1.5?A fun little problemWhat is the smallest integer $n$ greater than $1$ such that the root mean square of the first $n$ integers is an integer?on roots of an equationAn integer sequence with integer $k$ normsProof Verification: If $x$ is a nonnegative real number, then $big[sqrt[x]big] = big[sqrtxbig]$Digit after decimal point of radicalsProving that there does not exist an infinite descending sequence of naturals using minimal counterexamplenumber of different ways to represent a positive integer as a binomial coefficient













1












$begingroup$


Consider the equation $$f(x) =sqrtx^2 - x + 1$$



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: $$x in mathbbZ$$










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
    4 hours ago










  • $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
    4 hours ago










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbbZ$$
    $endgroup$
    – Diehardwalnut
    2 hours ago















1












$begingroup$


Consider the equation $$f(x) =sqrtx^2 - x + 1$$



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: $$x in mathbbZ$$










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
    4 hours ago










  • $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
    4 hours ago










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbbZ$$
    $endgroup$
    – Diehardwalnut
    2 hours ago













1












1








1


1



$begingroup$


Consider the equation $$f(x) =sqrtx^2 - x + 1$$



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: $$x in mathbbZ$$










share|cite|improve this question











$endgroup$




Consider the equation $$f(x) =sqrtx^2 - x + 1$$



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: $$x in mathbbZ$$







elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







Diehardwalnut

















asked 4 hours ago









DiehardwalnutDiehardwalnut

257110




257110







  • 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
    4 hours ago










  • $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
    4 hours ago










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbbZ$$
    $endgroup$
    – Diehardwalnut
    2 hours ago












  • 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
    4 hours ago










  • $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
    4 hours ago










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbbZ$$
    $endgroup$
    – Diehardwalnut
    2 hours ago







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
4 hours ago




$begingroup$
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
4 hours ago












$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
4 hours ago




$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt13)/2$, 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
4 hours ago












$begingroup$
Justed edited, yes I meant to say $$x in mathbbZ$$
$endgroup$
– Diehardwalnut
2 hours ago




$begingroup$
Justed edited, yes I meant to say $$x in mathbbZ$$
$endgroup$
– Diehardwalnut
2 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

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 $$x^2-x+1=(x-1)^2+colorbluex$$



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



Furthermore
beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
x^2-(x-1)^2&=colorblue2x-1tag2
endalign*



Can you end it now?




Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.







share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    Hint: let $y = sqrtx^2-x+1$. 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









    $endgroup$




















      0












      $begingroup$

      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

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        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 $$x^2-x+1=(x-1)^2+colorbluex$$



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



        Furthermore
        beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
        x^2-(x-1)^2&=colorblue2x-1tag2
        endalign*



        Can you end it now?




        Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.







        share|cite|improve this answer











        $endgroup$

















          3












          $begingroup$

          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 $$x^2-x+1=(x-1)^2+colorbluex$$



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



          Furthermore
          beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
          x^2-(x-1)^2&=colorblue2x-1tag2
          endalign*



          Can you end it now?




          Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.







          share|cite|improve this answer











          $endgroup$















            3












            3








            3





            $begingroup$

            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 $$x^2-x+1=(x-1)^2+colorbluex$$



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



            Furthermore
            beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
            x^2-(x-1)^2&=colorblue2x-1tag2
            endalign*



            Can you end it now?




            Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.







            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 $$x^2-x+1=(x-1)^2+colorbluex$$



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



            Furthermore
            beginalign*(x-1)^2-(x-2)^2&=colorblue2x-3tag1\
            x^2-(x-1)^2&=colorblue2x-1tag2
            endalign*



            Can you end it now?




            Hint: Observe, for instance, that $$mid;2x-3mid>mid x;mid text unless xin[1, 3]$$ $$mid;2x-1mid>mid x;mid text unless xin[frac13, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin1, 2, 3$.








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 4 hours ago









            Dr. MathvaDr. Mathva

            3,215630




            3,215630





















                2












                $begingroup$

                Hint: let $y = sqrtx^2-x+1$. 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: let $y = sqrtx^2-x+1$. 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












                    2








                    2





                    $begingroup$

                    Hint: let $y = sqrtx^2-x+1$. 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 $y = sqrtx^2-x+1$. 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 4 hours ago









                    ThéophileThéophile

                    20.4k13047




                    20.4k13047





















                        0












                        $begingroup$

                        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









                        $endgroup$

















                          0












                          $begingroup$

                          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









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            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 2 hours ago









                            J. W. TannerJ. W. Tanner

                            4,4691320




                            4,4691320



























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