If $x^2+ax+b+1=0$ ($a,binmathbb{Z}$) has integral roots, prove that $a^2+b^2$ is composite.












1












$begingroup$



If $x^2+ax+b+1=0$ ($a,binmathbb{Z}$) has integral roots, prove that
$a^2+b^2$ is composite.




Would someone please help me to solve the above question? I'm not able to understand how I should proceed.
Take $bne-1$.






number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $b=/=1$ ?
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 9:33










  • $begingroup$
    I meant not equal to -1. I don't know how to type that symbol.
    $endgroup$
    – Shashwat1337
    Jan 14 at 9:34






  • 1




    $begingroup$
    Compute the discriminant.
    $endgroup$
    – Wuestenfux
    Jan 14 at 9:35






  • 2




    $begingroup$
    Let $$x^2+ax+b+1=(x-r)(x-s)$$ so that $r,s$ are the two roots, then try to compute $a^2+b^2$ in terms of $r,s$.
    $endgroup$
    – Yong Hao Ng
    Jan 14 at 9:49












  • $begingroup$
    Thanks. I got it.
    $endgroup$
    – Shashwat1337
    Jan 14 at 9:54
















1












$begingroup$



If $x^2+ax+b+1=0$ ($a,binmathbb{Z}$) has integral roots, prove that
$a^2+b^2$ is composite.




Would someone please help me to solve the above question? I'm not able to understand how I should proceed.
Take $bne-1$.






number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $b=/=1$ ?
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 9:33










  • $begingroup$
    I meant not equal to -1. I don't know how to type that symbol.
    $endgroup$
    – Shashwat1337
    Jan 14 at 9:34






  • 1




    $begingroup$
    Compute the discriminant.
    $endgroup$
    – Wuestenfux
    Jan 14 at 9:35






  • 2




    $begingroup$
    Let $$x^2+ax+b+1=(x-r)(x-s)$$ so that $r,s$ are the two roots, then try to compute $a^2+b^2$ in terms of $r,s$.
    $endgroup$
    – Yong Hao Ng
    Jan 14 at 9:49












  • $begingroup$
    Thanks. I got it.
    $endgroup$
    – Shashwat1337
    Jan 14 at 9:54














1












1








1





$begingroup$



If $x^2+ax+b+1=0$ ($a,binmathbb{Z}$) has integral roots, prove that
$a^2+b^2$ is composite.




Would someone please help me to solve the above question? I'm not able to understand how I should proceed.
Take $bne-1$.






number-theory







share|cite|improve this question











$endgroup$





If $x^2+ax+b+1=0$ ($a,binmathbb{Z}$) has integral roots, prove that
$a^2+b^2$ is composite.




Would someone please help me to solve the above question? I'm not able to understand how I should proceed.
Take $bne-1$.






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 Jan 14 at 10:25









rtybase

11k21533




11k21533










asked Jan 14 at 9:31









Shashwat1337Shashwat1337

789




789












  • $begingroup$
    What is $b=/=1$ ?
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 9:33










  • $begingroup$
    I meant not equal to -1. I don't know how to type that symbol.
    $endgroup$
    – Shashwat1337
    Jan 14 at 9:34






  • 1




    $begingroup$
    Compute the discriminant.
    $endgroup$
    – Wuestenfux
    Jan 14 at 9:35






  • 2




    $begingroup$
    Let $$x^2+ax+b+1=(x-r)(x-s)$$ so that $r,s$ are the two roots, then try to compute $a^2+b^2$ in terms of $r,s$.
    $endgroup$
    – Yong Hao Ng
    Jan 14 at 9:49












  • $begingroup$
    Thanks. I got it.
    $endgroup$
    – Shashwat1337
    Jan 14 at 9:54


















  • $begingroup$
    What is $b=/=1$ ?
    $endgroup$
    – Sauhard Sharma
    Jan 14 at 9:33










  • $begingroup$
    I meant not equal to -1. I don't know how to type that symbol.
    $endgroup$
    – Shashwat1337
    Jan 14 at 9:34






  • 1




    $begingroup$
    Compute the discriminant.
    $endgroup$
    – Wuestenfux
    Jan 14 at 9:35






  • 2




    $begingroup$
    Let $$x^2+ax+b+1=(x-r)(x-s)$$ so that $r,s$ are the two roots, then try to compute $a^2+b^2$ in terms of $r,s$.
    $endgroup$
    – Yong Hao Ng
    Jan 14 at 9:49












  • $begingroup$
    Thanks. I got it.
    $endgroup$
    – Shashwat1337
    Jan 14 at 9:54
















$begingroup$
What is $b=/=1$ ?
$endgroup$
– Sauhard Sharma
Jan 14 at 9:33




$begingroup$
What is $b=/=1$ ?
$endgroup$
– Sauhard Sharma
Jan 14 at 9:33












$begingroup$
I meant not equal to -1. I don't know how to type that symbol.
$endgroup$
– Shashwat1337
Jan 14 at 9:34




$begingroup$
I meant not equal to -1. I don't know how to type that symbol.
$endgroup$
– Shashwat1337
Jan 14 at 9:34




1




1




$begingroup$
Compute the discriminant.
$endgroup$
– Wuestenfux
Jan 14 at 9:35




$begingroup$
Compute the discriminant.
$endgroup$
– Wuestenfux
Jan 14 at 9:35




2




2




$begingroup$
Let $$x^2+ax+b+1=(x-r)(x-s)$$ so that $r,s$ are the two roots, then try to compute $a^2+b^2$ in terms of $r,s$.
$endgroup$
– Yong Hao Ng
Jan 14 at 9:49






$begingroup$
Let $$x^2+ax+b+1=(x-r)(x-s)$$ so that $r,s$ are the two roots, then try to compute $a^2+b^2$ in terms of $r,s$.
$endgroup$
– Yong Hao Ng
Jan 14 at 9:49














$begingroup$
Thanks. I got it.
$endgroup$
– Shashwat1337
Jan 14 at 9:54




$begingroup$
Thanks. I got it.
$endgroup$
– Shashwat1337
Jan 14 at 9:54










1 Answer
1






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oldest

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7












$begingroup$

From Vieta's
$$x_1 + x_2 = -a$$
$$x_1 cdot x_2 = b+1$$
or
$$a^2+b^2=left(x_1+x_2right)^2+left(x_1cdot x_2-1right)^2=\
x_1^2+x_2^2+2x_1x_2+x_1^2x_2^2-2x_1x_2+1=\
x_1^2+x_2^2+x_1^2x_2^2+1=\
left(x_1^2+1right)left(x_2^2+1right)$$
Since $bne-1$, then none of $x_1,x_2$ is $0$ and the result follows.






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    7












    $begingroup$

    From Vieta's
    $$x_1 + x_2 = -a$$
    $$x_1 cdot x_2 = b+1$$
    or
    $$a^2+b^2=left(x_1+x_2right)^2+left(x_1cdot x_2-1right)^2=\
    x_1^2+x_2^2+2x_1x_2+x_1^2x_2^2-2x_1x_2+1=\
    x_1^2+x_2^2+x_1^2x_2^2+1=\
    left(x_1^2+1right)left(x_2^2+1right)$$
    Since $bne-1$, then none of $x_1,x_2$ is $0$ and the result follows.






    share|cite|improve this answer











    $endgroup$


















      7












      $begingroup$

      From Vieta's
      $$x_1 + x_2 = -a$$
      $$x_1 cdot x_2 = b+1$$
      or
      $$a^2+b^2=left(x_1+x_2right)^2+left(x_1cdot x_2-1right)^2=\
      x_1^2+x_2^2+2x_1x_2+x_1^2x_2^2-2x_1x_2+1=\
      x_1^2+x_2^2+x_1^2x_2^2+1=\
      left(x_1^2+1right)left(x_2^2+1right)$$
      Since $bne-1$, then none of $x_1,x_2$ is $0$ and the result follows.






      share|cite|improve this answer











      $endgroup$
















        7












        7








        7





        $begingroup$

        From Vieta's
        $$x_1 + x_2 = -a$$
        $$x_1 cdot x_2 = b+1$$
        or
        $$a^2+b^2=left(x_1+x_2right)^2+left(x_1cdot x_2-1right)^2=\
        x_1^2+x_2^2+2x_1x_2+x_1^2x_2^2-2x_1x_2+1=\
        x_1^2+x_2^2+x_1^2x_2^2+1=\
        left(x_1^2+1right)left(x_2^2+1right)$$
        Since $bne-1$, then none of $x_1,x_2$ is $0$ and the result follows.






        share|cite|improve this answer











        $endgroup$



        From Vieta's
        $$x_1 + x_2 = -a$$
        $$x_1 cdot x_2 = b+1$$
        or
        $$a^2+b^2=left(x_1+x_2right)^2+left(x_1cdot x_2-1right)^2=\
        x_1^2+x_2^2+2x_1x_2+x_1^2x_2^2-2x_1x_2+1=\
        x_1^2+x_2^2+x_1^2x_2^2+1=\
        left(x_1^2+1right)left(x_2^2+1right)$$
        Since $bne-1$, then none of $x_1,x_2$ is $0$ and the result follows.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 14 at 10:22

























        answered Jan 14 at 10:13









        rtybasertybase

        11k21533




        11k21533






























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