n-th order polynomial with all roots where all coefficients are 1 or -1, highest order of n?












7












$begingroup$


I have polynomial $f(x） = sum_{i=0}^n a_i x^i $, where $a_i = pm 1$. All roots of $f(x)$ are real. What's the highest order of $n$? Note that the roots are real but can be irrational. Even if there are duplicate roots, that's fine






number-theory polynomials







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$endgroup$












  • $begingroup$
    I assume you are not counting multiplicity of the roots (i.e. the n roots are all distinct).
    $endgroup$
    – D.B.
    Dec 25 '18 at 4:02










  • $begingroup$
    @D.B. yes, even if there are duplicate roots, that's fine
    $endgroup$
    – CodeNoob
    Dec 25 '18 at 4:04






  • 1




    $begingroup$
    An example with $n=3$ is $x^3-x^2-x+1 = (x-1)^2(x+1)$. There don't seem to be any for $n=4, 5$, $6$ or $7$.
    $endgroup$
    – Robert Israel
    Dec 25 '18 at 4:15












  • $begingroup$
    @RobertIsrael that's right. We can prove all roots have abs value between (0.5,2) as well.
    $endgroup$
    – CodeNoob
    Dec 25 '18 at 4:40










  • $begingroup$
    Descartes’ Rule of Signs might be helpful here.
    $endgroup$
    – Clayton
    Dec 25 '18 at 4:49
















7












$begingroup$


I have polynomial $f(x） = sum_{i=0}^n a_i x^i $, where $a_i = pm 1$. All roots of $f(x)$ are real. What's the highest order of $n$? Note that the roots are real but can be irrational. Even if there are duplicate roots, that's fine






number-theory polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    I assume you are not counting multiplicity of the roots (i.e. the n roots are all distinct).
    $endgroup$
    – D.B.
    Dec 25 '18 at 4:02










  • $begingroup$
    @D.B. yes, even if there are duplicate roots, that's fine
    $endgroup$
    – CodeNoob
    Dec 25 '18 at 4:04






  • 1




    $begingroup$
    An example with $n=3$ is $x^3-x^2-x+1 = (x-1)^2(x+1)$. There don't seem to be any for $n=4, 5$, $6$ or $7$.
    $endgroup$
    – Robert Israel
    Dec 25 '18 at 4:15












  • $begingroup$
    @RobertIsrael that's right. We can prove all roots have abs value between (0.5,2) as well.
    $endgroup$
    – CodeNoob
    Dec 25 '18 at 4:40










  • $begingroup$
    Descartes’ Rule of Signs might be helpful here.
    $endgroup$
    – Clayton
    Dec 25 '18 at 4:49














7












7








7


2



$begingroup$


I have polynomial $f(x） = sum_{i=0}^n a_i x^i $, where $a_i = pm 1$. All roots of $f(x)$ are real. What's the highest order of $n$? Note that the roots are real but can be irrational. Even if there are duplicate roots, that's fine






number-theory polynomials







share|cite|improve this question











$endgroup$




I have polynomial $f(x） = sum_{i=0}^n a_i x^i $, where $a_i = pm 1$. All roots of $f(x)$ are real. What's the highest order of $n$? Note that the roots are real but can be irrational. Even if there are duplicate roots, that's fine






number-theory polynomials




number-theory polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 14:37







CodeNoob

















asked Dec 25 '18 at 3:45









CodeNoobCodeNoob

23019




23019












  • $begingroup$
    I assume you are not counting multiplicity of the roots (i.e. the n roots are all distinct).
    $endgroup$
    – D.B.
    Dec 25 '18 at 4:02










  • $begingroup$
    @D.B. yes, even if there are duplicate roots, that's fine
    $endgroup$
    – CodeNoob
    Dec 25 '18 at 4:04






  • 1




    $begingroup$
    An example with $n=3$ is $x^3-x^2-x+1 = (x-1)^2(x+1)$. There don't seem to be any for $n=4, 5$, $6$ or $7$.
    $endgroup$
    – Robert Israel
    Dec 25 '18 at 4:15












  • $begingroup$
    @RobertIsrael that's right. We can prove all roots have abs value between (0.5,2) as well.
    $endgroup$
    – CodeNoob
    Dec 25 '18 at 4:40










  • $begingroup$
    Descartes’ Rule of Signs might be helpful here.
    $endgroup$
    – Clayton
    Dec 25 '18 at 4:49


















  • $begingroup$
    I assume you are not counting multiplicity of the roots (i.e. the n roots are all distinct).
    $endgroup$
    – D.B.
    Dec 25 '18 at 4:02










  • $begingroup$
    @D.B. yes, even if there are duplicate roots, that's fine
    $endgroup$
    – CodeNoob
    Dec 25 '18 at 4:04






  • 1




    $begingroup$
    An example with $n=3$ is $x^3-x^2-x+1 = (x-1)^2(x+1)$. There don't seem to be any for $n=4, 5$, $6$ or $7$.
    $endgroup$
    – Robert Israel
    Dec 25 '18 at 4:15












  • $begingroup$
    @RobertIsrael that's right. We can prove all roots have abs value between (0.5,2) as well.
    $endgroup$
    – CodeNoob
    Dec 25 '18 at 4:40










  • $begingroup$
    Descartes’ Rule of Signs might be helpful here.
    $endgroup$
    – Clayton
    Dec 25 '18 at 4:49
















$begingroup$
I assume you are not counting multiplicity of the roots (i.e. the n roots are all distinct).
$endgroup$
– D.B.
Dec 25 '18 at 4:02




$begingroup$
I assume you are not counting multiplicity of the roots (i.e. the n roots are all distinct).
$endgroup$
– D.B.
Dec 25 '18 at 4:02












$begingroup$
@D.B. yes, even if there are duplicate roots, that's fine
$endgroup$
– CodeNoob
Dec 25 '18 at 4:04




$begingroup$
@D.B. yes, even if there are duplicate roots, that's fine
$endgroup$
– CodeNoob
Dec 25 '18 at 4:04




1




1




$begingroup$
An example with $n=3$ is $x^3-x^2-x+1 = (x-1)^2(x+1)$. There don't seem to be any for $n=4, 5$, $6$ or $7$.
$endgroup$
– Robert Israel
Dec 25 '18 at 4:15






$begingroup$
An example with $n=3$ is $x^3-x^2-x+1 = (x-1)^2(x+1)$. There don't seem to be any for $n=4, 5$, $6$ or $7$.
$endgroup$
– Robert Israel
Dec 25 '18 at 4:15














$begingroup$
@RobertIsrael that's right. We can prove all roots have abs value between (0.5,2) as well.
$endgroup$
– CodeNoob
Dec 25 '18 at 4:40




$begingroup$
@RobertIsrael that's right. We can prove all roots have abs value between (0.5,2) as well.
$endgroup$
– CodeNoob
Dec 25 '18 at 4:40












$begingroup$
Descartes’ Rule of Signs might be helpful here.
$endgroup$
– Clayton
Dec 25 '18 at 4:49




$begingroup$
Descartes’ Rule of Signs might be helpful here.
$endgroup$
– Clayton
Dec 25 '18 at 4:49










1 Answer
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$begingroup$

Vieta's Formulas are the key to this problem. Let $r_1, cdots, r_n$ be the roots. Then define
begin{align}
A &= sum_{i=1}^n r_i \
B &= sum_{1 le i_2 < i_2 le n} r_{i_1}r_{i_2} \
C &= prod_{i=1}^n r_i .
end{align}
By Vieta's Formulas, we know that $A, B, C in {pm 1}.$ Now
$$sum_{i=1}^n r_i^2 = A^2 - 2B = 1-2B ge 0 implies B = -1.$$
But then by AM-GM, we have
$$frac{3}n = frac{1}n sum_{i=1}^n r_i^2 ge left(C^2 right)^{1/n} = 1 $$
which cannot happen for $n ge 4$.






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    1 Answer
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    8












    $begingroup$

    Vieta's Formulas are the key to this problem. Let $r_1, cdots, r_n$ be the roots. Then define
    begin{align}
    A &= sum_{i=1}^n r_i \
    B &= sum_{1 le i_2 < i_2 le n} r_{i_1}r_{i_2} \
    C &= prod_{i=1}^n r_i .
    end{align}
    By Vieta's Formulas, we know that $A, B, C in {pm 1}.$ Now
    $$sum_{i=1}^n r_i^2 = A^2 - 2B = 1-2B ge 0 implies B = -1.$$
    But then by AM-GM, we have
    $$frac{3}n = frac{1}n sum_{i=1}^n r_i^2 ge left(C^2 right)^{1/n} = 1 $$
    which cannot happen for $n ge 4$.






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      Vieta's Formulas are the key to this problem. Let $r_1, cdots, r_n$ be the roots. Then define
      begin{align}
      A &= sum_{i=1}^n r_i \
      B &= sum_{1 le i_2 < i_2 le n} r_{i_1}r_{i_2} \
      C &= prod_{i=1}^n r_i .
      end{align}
      By Vieta's Formulas, we know that $A, B, C in {pm 1}.$ Now
      $$sum_{i=1}^n r_i^2 = A^2 - 2B = 1-2B ge 0 implies B = -1.$$
      But then by AM-GM, we have
      $$frac{3}n = frac{1}n sum_{i=1}^n r_i^2 ge left(C^2 right)^{1/n} = 1 $$
      which cannot happen for $n ge 4$.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        Vieta's Formulas are the key to this problem. Let $r_1, cdots, r_n$ be the roots. Then define
        begin{align}
        A &= sum_{i=1}^n r_i \
        B &= sum_{1 le i_2 < i_2 le n} r_{i_1}r_{i_2} \
        C &= prod_{i=1}^n r_i .
        end{align}
        By Vieta's Formulas, we know that $A, B, C in {pm 1}.$ Now
        $$sum_{i=1}^n r_i^2 = A^2 - 2B = 1-2B ge 0 implies B = -1.$$
        But then by AM-GM, we have
        $$frac{3}n = frac{1}n sum_{i=1}^n r_i^2 ge left(C^2 right)^{1/n} = 1 $$
        which cannot happen for $n ge 4$.






        share|cite|improve this answer









        $endgroup$



        Vieta's Formulas are the key to this problem. Let $r_1, cdots, r_n$ be the roots. Then define
        begin{align}
        A &= sum_{i=1}^n r_i \
        B &= sum_{1 le i_2 < i_2 le n} r_{i_1}r_{i_2} \
        C &= prod_{i=1}^n r_i .
        end{align}
        By Vieta's Formulas, we know that $A, B, C in {pm 1}.$ Now
        $$sum_{i=1}^n r_i^2 = A^2 - 2B = 1-2B ge 0 implies B = -1.$$
        But then by AM-GM, we have
        $$frac{3}n = frac{1}n sum_{i=1}^n r_i^2 ge left(C^2 right)^{1/n} = 1 $$
        which cannot happen for $n ge 4$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 5:15









        Sandeep SilwalSandeep Silwal

        5,91811337




        5,91811337






























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