Question about number of perfect squares between two perfect cubes












-2












$begingroup$


Prove that for a natural number $k≥2$ , there are a minimum of two perfect squares in the interval $(k^3,(k+1)^3)$



I tried induction , supposing that there are $m^2<n^2$ in the interval , and to prove that there exists two perfect squares $q^2<r^2$ in $((k+1)^3,(k+2)^3))$.






number-theory







share|cite|improve this question











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  • 1




    $begingroup$
    Welcome to stackexchange. If you edit the question to show us what you tried and where you are stuck we may be able to help.
    $endgroup$
    – Ethan Bolker
    Dec 14 '18 at 18:53










  • $begingroup$
    As a hint to get started: can you prove that there is at least one perfect square in the interval?
    $endgroup$
    – lulu
    Dec 14 '18 at 18:59










  • $begingroup$
    Yes , I think it's $([sqrt{k^3}]^2)$
    $endgroup$
    – reducere cs
    Dec 14 '18 at 19:08












  • $begingroup$
    Hint: for $kge 2$, $(k+1)^{3/2}-k^{3/2}=frac{3}{2}sqrt{k}>2$.
    $endgroup$
    – J.G.
    Dec 14 '18 at 19:10










  • $begingroup$
    So there are at least two integers between $(k+1)^frac{3}{2} - k^frac{3}{2}$ or
    $endgroup$
    – reducere cs
    Dec 14 '18 at 19:15


















-2












$begingroup$


Prove that for a natural number $k≥2$ , there are a minimum of two perfect squares in the interval $(k^3,(k+1)^3)$



I tried induction , supposing that there are $m^2<n^2$ in the interval , and to prove that there exists two perfect squares $q^2<r^2$ in $((k+1)^3,(k+2)^3))$.






number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to stackexchange. If you edit the question to show us what you tried and where you are stuck we may be able to help.
    $endgroup$
    – Ethan Bolker
    Dec 14 '18 at 18:53










  • $begingroup$
    As a hint to get started: can you prove that there is at least one perfect square in the interval?
    $endgroup$
    – lulu
    Dec 14 '18 at 18:59










  • $begingroup$
    Yes , I think it's $([sqrt{k^3}]^2)$
    $endgroup$
    – reducere cs
    Dec 14 '18 at 19:08












  • $begingroup$
    Hint: for $kge 2$, $(k+1)^{3/2}-k^{3/2}=frac{3}{2}sqrt{k}>2$.
    $endgroup$
    – J.G.
    Dec 14 '18 at 19:10










  • $begingroup$
    So there are at least two integers between $(k+1)^frac{3}{2} - k^frac{3}{2}$ or
    $endgroup$
    – reducere cs
    Dec 14 '18 at 19:15
















-2












-2








-2





$begingroup$


Prove that for a natural number $k≥2$ , there are a minimum of two perfect squares in the interval $(k^3,(k+1)^3)$



I tried induction , supposing that there are $m^2<n^2$ in the interval , and to prove that there exists two perfect squares $q^2<r^2$ in $((k+1)^3,(k+2)^3))$.






number-theory







share|cite|improve this question











$endgroup$




Prove that for a natural number $k≥2$ , there are a minimum of two perfect squares in the interval $(k^3,(k+1)^3)$



I tried induction , supposing that there are $m^2<n^2$ in the interval , and to prove that there exists two perfect squares $q^2<r^2$ in $((k+1)^3,(k+2)^3))$.






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 18:56







reducere cs

















asked Dec 14 '18 at 18:52









reducere csreducere cs

112




112








  • 1




    $begingroup$
    Welcome to stackexchange. If you edit the question to show us what you tried and where you are stuck we may be able to help.
    $endgroup$
    – Ethan Bolker
    Dec 14 '18 at 18:53










  • $begingroup$
    As a hint to get started: can you prove that there is at least one perfect square in the interval?
    $endgroup$
    – lulu
    Dec 14 '18 at 18:59










  • $begingroup$
    Yes , I think it's $([sqrt{k^3}]^2)$
    $endgroup$
    – reducere cs
    Dec 14 '18 at 19:08












  • $begingroup$
    Hint: for $kge 2$, $(k+1)^{3/2}-k^{3/2}=frac{3}{2}sqrt{k}>2$.
    $endgroup$
    – J.G.
    Dec 14 '18 at 19:10










  • $begingroup$
    So there are at least two integers between $(k+1)^frac{3}{2} - k^frac{3}{2}$ or
    $endgroup$
    – reducere cs
    Dec 14 '18 at 19:15
















  • 1




    $begingroup$
    Welcome to stackexchange. If you edit the question to show us what you tried and where you are stuck we may be able to help.
    $endgroup$
    – Ethan Bolker
    Dec 14 '18 at 18:53










  • $begingroup$
    As a hint to get started: can you prove that there is at least one perfect square in the interval?
    $endgroup$
    – lulu
    Dec 14 '18 at 18:59










  • $begingroup$
    Yes , I think it's $([sqrt{k^3}]^2)$
    $endgroup$
    – reducere cs
    Dec 14 '18 at 19:08












  • $begingroup$
    Hint: for $kge 2$, $(k+1)^{3/2}-k^{3/2}=frac{3}{2}sqrt{k}>2$.
    $endgroup$
    – J.G.
    Dec 14 '18 at 19:10










  • $begingroup$
    So there are at least two integers between $(k+1)^frac{3}{2} - k^frac{3}{2}$ or
    $endgroup$
    – reducere cs
    Dec 14 '18 at 19:15










1




1




$begingroup$
Welcome to stackexchange. If you edit the question to show us what you tried and where you are stuck we may be able to help.
$endgroup$
– Ethan Bolker
Dec 14 '18 at 18:53




$begingroup$
Welcome to stackexchange. If you edit the question to show us what you tried and where you are stuck we may be able to help.
$endgroup$
– Ethan Bolker
Dec 14 '18 at 18:53












$begingroup$
As a hint to get started: can you prove that there is at least one perfect square in the interval?
$endgroup$
– lulu
Dec 14 '18 at 18:59




$begingroup$
As a hint to get started: can you prove that there is at least one perfect square in the interval?
$endgroup$
– lulu
Dec 14 '18 at 18:59












$begingroup$
Yes , I think it's $([sqrt{k^3}]^2)$
$endgroup$
– reducere cs
Dec 14 '18 at 19:08






$begingroup$
Yes , I think it's $([sqrt{k^3}]^2)$
$endgroup$
– reducere cs
Dec 14 '18 at 19:08














$begingroup$
Hint: for $kge 2$, $(k+1)^{3/2}-k^{3/2}=frac{3}{2}sqrt{k}>2$.
$endgroup$
– J.G.
Dec 14 '18 at 19:10




$begingroup$
Hint: for $kge 2$, $(k+1)^{3/2}-k^{3/2}=frac{3}{2}sqrt{k}>2$.
$endgroup$
– J.G.
Dec 14 '18 at 19:10












$begingroup$
So there are at least two integers between $(k+1)^frac{3}{2} - k^frac{3}{2}$ or
$endgroup$
– reducere cs
Dec 14 '18 at 19:15






$begingroup$
So there are at least two integers between $(k+1)^frac{3}{2} - k^frac{3}{2}$ or
$endgroup$
– reducere cs
Dec 14 '18 at 19:15












1 Answer
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$begingroup$

Let $m$ be the largest integer such that $m^2le k^3$



$mle k^{frac 32}$



if $k>1$



$4m le 4k^{frac 32} < 3k^2$ and $4<3k+1$



$k^3<(m+1)^2<(m+2)^2 = m^2 + 4m + 4 < k^3 + 3k^2 + 3k + 1 = (k+1)^3$






share|cite|improve this answer









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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

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    active

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    0












    $begingroup$

    Let $m$ be the largest integer such that $m^2le k^3$



    $mle k^{frac 32}$



    if $k>1$



    $4m le 4k^{frac 32} < 3k^2$ and $4<3k+1$



    $k^3<(m+1)^2<(m+2)^2 = m^2 + 4m + 4 < k^3 + 3k^2 + 3k + 1 = (k+1)^3$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let $m$ be the largest integer such that $m^2le k^3$



      $mle k^{frac 32}$



      if $k>1$



      $4m le 4k^{frac 32} < 3k^2$ and $4<3k+1$



      $k^3<(m+1)^2<(m+2)^2 = m^2 + 4m + 4 < k^3 + 3k^2 + 3k + 1 = (k+1)^3$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $m$ be the largest integer such that $m^2le k^3$



        $mle k^{frac 32}$



        if $k>1$



        $4m le 4k^{frac 32} < 3k^2$ and $4<3k+1$



        $k^3<(m+1)^2<(m+2)^2 = m^2 + 4m + 4 < k^3 + 3k^2 + 3k + 1 = (k+1)^3$






        share|cite|improve this answer









        $endgroup$



        Let $m$ be the largest integer such that $m^2le k^3$



        $mle k^{frac 32}$



        if $k>1$



        $4m le 4k^{frac 32} < 3k^2$ and $4<3k+1$



        $k^3<(m+1)^2<(m+2)^2 = m^2 + 4m + 4 < k^3 + 3k^2 + 3k + 1 = (k+1)^3$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 19:22









        Doug MDoug M

        44.9k31854




        44.9k31854






























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