Find all $a, b in mathbb R$, ($bne0)$, such that the roots of $x^2+ax+a=b$ and $x^2+ax+a=-b$ are 4...












0












$begingroup$


Find all $a, b in Bbb R$, ($bne0)$, such that the roots of
$$x^2+ax+a=b$$ $$x^2+ax+a=-b$$
are 4 consecutive numbers.



We have: $$x^2+ax+a-b=0$$ $$x^2+ax+a+b=0$$
$x_1, x_2$ - roots of first equation; $x_3, x_4$ - roots of second equation.



By logic I wrote: $x_2<x_4<x_3<x_1$ (Is there a way to prove this?)



Using Vieta's formulas: $$x_1+x_2+x_3+x_4=2a$$ and because they are consecutive numbers we have $$x=frac{a-3}2$$



$$frac{a-3}2=x_2<frac{a-1}2=x_4<frac{a+1}2=x_3<frac{a+3}2=x_1$$



Using Vieta's formulas again: $$frac{a^2-9}4=a-b,quad frac{a^2-1}4=a+b$$



$$implies b=-1$$
$$implies a_{1/2}=2pmfrac{sqrt{58}}2,quad a_3=5,quad a_4=-1$$



My question is are these all the solutions? Or are there more?










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  • $begingroup$
    You need to try the other products. Try $x_2cdot x_4$ for example.
    $endgroup$
    – Don Thousand
    Dec 4 '18 at 17:40










  • $begingroup$
    Although you already received answers, I think your question should be clarified: (1) 4 consecutive numbers: Do you mean integers ´$n, n+1, n+2, n+3$? (2) What are the $x_i$? Certainly the solutions of the two equations, but which $x_i$ of which equation? (3) Typo: $x = ...$.
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:21










  • $begingroup$
    I edited for $x_i$. Can consecutive numbers not be integers? (I got confused) @PaulFrost
    $endgroup$
    – Pero
    Dec 4 '18 at 18:27










  • $begingroup$
    I do not think so, but in your comment to Math Lover' s answer ending with "In that case, $a$ and $b$ must be integers" you said "$a,b$ can be real numbers". This confused me. Perhaps you edit once more and write "4 consecutive integers".
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:35










  • $begingroup$
    The problem / task says: $a, b in Bbb R$, ... the roots of the equations are 4 consecutive numbers. I guess what Math Lover was saying is that the roots won't be integers if $a, b$ are not integers.
    $endgroup$
    – Pero
    Dec 4 '18 at 18:38
















0












$begingroup$


Find all $a, b in Bbb R$, ($bne0)$, such that the roots of
$$x^2+ax+a=b$$ $$x^2+ax+a=-b$$
are 4 consecutive numbers.



We have: $$x^2+ax+a-b=0$$ $$x^2+ax+a+b=0$$
$x_1, x_2$ - roots of first equation; $x_3, x_4$ - roots of second equation.



By logic I wrote: $x_2<x_4<x_3<x_1$ (Is there a way to prove this?)



Using Vieta's formulas: $$x_1+x_2+x_3+x_4=2a$$ and because they are consecutive numbers we have $$x=frac{a-3}2$$



$$frac{a-3}2=x_2<frac{a-1}2=x_4<frac{a+1}2=x_3<frac{a+3}2=x_1$$



Using Vieta's formulas again: $$frac{a^2-9}4=a-b,quad frac{a^2-1}4=a+b$$



$$implies b=-1$$
$$implies a_{1/2}=2pmfrac{sqrt{58}}2,quad a_3=5,quad a_4=-1$$



My question is are these all the solutions? Or are there more?










share|cite|improve this question











$endgroup$












  • $begingroup$
    You need to try the other products. Try $x_2cdot x_4$ for example.
    $endgroup$
    – Don Thousand
    Dec 4 '18 at 17:40










  • $begingroup$
    Although you already received answers, I think your question should be clarified: (1) 4 consecutive numbers: Do you mean integers ´$n, n+1, n+2, n+3$? (2) What are the $x_i$? Certainly the solutions of the two equations, but which $x_i$ of which equation? (3) Typo: $x = ...$.
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:21










  • $begingroup$
    I edited for $x_i$. Can consecutive numbers not be integers? (I got confused) @PaulFrost
    $endgroup$
    – Pero
    Dec 4 '18 at 18:27










  • $begingroup$
    I do not think so, but in your comment to Math Lover' s answer ending with "In that case, $a$ and $b$ must be integers" you said "$a,b$ can be real numbers". This confused me. Perhaps you edit once more and write "4 consecutive integers".
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:35










  • $begingroup$
    The problem / task says: $a, b in Bbb R$, ... the roots of the equations are 4 consecutive numbers. I guess what Math Lover was saying is that the roots won't be integers if $a, b$ are not integers.
    $endgroup$
    – Pero
    Dec 4 '18 at 18:38














0












0








0





$begingroup$


Find all $a, b in Bbb R$, ($bne0)$, such that the roots of
$$x^2+ax+a=b$$ $$x^2+ax+a=-b$$
are 4 consecutive numbers.



We have: $$x^2+ax+a-b=0$$ $$x^2+ax+a+b=0$$
$x_1, x_2$ - roots of first equation; $x_3, x_4$ - roots of second equation.



By logic I wrote: $x_2<x_4<x_3<x_1$ (Is there a way to prove this?)



Using Vieta's formulas: $$x_1+x_2+x_3+x_4=2a$$ and because they are consecutive numbers we have $$x=frac{a-3}2$$



$$frac{a-3}2=x_2<frac{a-1}2=x_4<frac{a+1}2=x_3<frac{a+3}2=x_1$$



Using Vieta's formulas again: $$frac{a^2-9}4=a-b,quad frac{a^2-1}4=a+b$$



$$implies b=-1$$
$$implies a_{1/2}=2pmfrac{sqrt{58}}2,quad a_3=5,quad a_4=-1$$



My question is are these all the solutions? Or are there more?










share|cite|improve this question











$endgroup$




Find all $a, b in Bbb R$, ($bne0)$, such that the roots of
$$x^2+ax+a=b$$ $$x^2+ax+a=-b$$
are 4 consecutive numbers.



We have: $$x^2+ax+a-b=0$$ $$x^2+ax+a+b=0$$
$x_1, x_2$ - roots of first equation; $x_3, x_4$ - roots of second equation.



By logic I wrote: $x_2<x_4<x_3<x_1$ (Is there a way to prove this?)



Using Vieta's formulas: $$x_1+x_2+x_3+x_4=2a$$ and because they are consecutive numbers we have $$x=frac{a-3}2$$



$$frac{a-3}2=x_2<frac{a-1}2=x_4<frac{a+1}2=x_3<frac{a+3}2=x_1$$



Using Vieta's formulas again: $$frac{a^2-9}4=a-b,quad frac{a^2-1}4=a+b$$



$$implies b=-1$$
$$implies a_{1/2}=2pmfrac{sqrt{58}}2,quad a_3=5,quad a_4=-1$$



My question is are these all the solutions? Or are there more?







algebra-precalculus quadratics






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edited Dec 4 '18 at 18:25







Pero

















asked Dec 4 '18 at 17:19









PeroPero

1207




1207












  • $begingroup$
    You need to try the other products. Try $x_2cdot x_4$ for example.
    $endgroup$
    – Don Thousand
    Dec 4 '18 at 17:40










  • $begingroup$
    Although you already received answers, I think your question should be clarified: (1) 4 consecutive numbers: Do you mean integers ´$n, n+1, n+2, n+3$? (2) What are the $x_i$? Certainly the solutions of the two equations, but which $x_i$ of which equation? (3) Typo: $x = ...$.
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:21










  • $begingroup$
    I edited for $x_i$. Can consecutive numbers not be integers? (I got confused) @PaulFrost
    $endgroup$
    – Pero
    Dec 4 '18 at 18:27










  • $begingroup$
    I do not think so, but in your comment to Math Lover' s answer ending with "In that case, $a$ and $b$ must be integers" you said "$a,b$ can be real numbers". This confused me. Perhaps you edit once more and write "4 consecutive integers".
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:35










  • $begingroup$
    The problem / task says: $a, b in Bbb R$, ... the roots of the equations are 4 consecutive numbers. I guess what Math Lover was saying is that the roots won't be integers if $a, b$ are not integers.
    $endgroup$
    – Pero
    Dec 4 '18 at 18:38


















  • $begingroup$
    You need to try the other products. Try $x_2cdot x_4$ for example.
    $endgroup$
    – Don Thousand
    Dec 4 '18 at 17:40










  • $begingroup$
    Although you already received answers, I think your question should be clarified: (1) 4 consecutive numbers: Do you mean integers ´$n, n+1, n+2, n+3$? (2) What are the $x_i$? Certainly the solutions of the two equations, but which $x_i$ of which equation? (3) Typo: $x = ...$.
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:21










  • $begingroup$
    I edited for $x_i$. Can consecutive numbers not be integers? (I got confused) @PaulFrost
    $endgroup$
    – Pero
    Dec 4 '18 at 18:27










  • $begingroup$
    I do not think so, but in your comment to Math Lover' s answer ending with "In that case, $a$ and $b$ must be integers" you said "$a,b$ can be real numbers". This confused me. Perhaps you edit once more and write "4 consecutive integers".
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:35










  • $begingroup$
    The problem / task says: $a, b in Bbb R$, ... the roots of the equations are 4 consecutive numbers. I guess what Math Lover was saying is that the roots won't be integers if $a, b$ are not integers.
    $endgroup$
    – Pero
    Dec 4 '18 at 18:38
















$begingroup$
You need to try the other products. Try $x_2cdot x_4$ for example.
$endgroup$
– Don Thousand
Dec 4 '18 at 17:40




$begingroup$
You need to try the other products. Try $x_2cdot x_4$ for example.
$endgroup$
– Don Thousand
Dec 4 '18 at 17:40












$begingroup$
Although you already received answers, I think your question should be clarified: (1) 4 consecutive numbers: Do you mean integers ´$n, n+1, n+2, n+3$? (2) What are the $x_i$? Certainly the solutions of the two equations, but which $x_i$ of which equation? (3) Typo: $x = ...$.
$endgroup$
– Paul Frost
Dec 4 '18 at 18:21




$begingroup$
Although you already received answers, I think your question should be clarified: (1) 4 consecutive numbers: Do you mean integers ´$n, n+1, n+2, n+3$? (2) What are the $x_i$? Certainly the solutions of the two equations, but which $x_i$ of which equation? (3) Typo: $x = ...$.
$endgroup$
– Paul Frost
Dec 4 '18 at 18:21












$begingroup$
I edited for $x_i$. Can consecutive numbers not be integers? (I got confused) @PaulFrost
$endgroup$
– Pero
Dec 4 '18 at 18:27




$begingroup$
I edited for $x_i$. Can consecutive numbers not be integers? (I got confused) @PaulFrost
$endgroup$
– Pero
Dec 4 '18 at 18:27












$begingroup$
I do not think so, but in your comment to Math Lover' s answer ending with "In that case, $a$ and $b$ must be integers" you said "$a,b$ can be real numbers". This confused me. Perhaps you edit once more and write "4 consecutive integers".
$endgroup$
– Paul Frost
Dec 4 '18 at 18:35




$begingroup$
I do not think so, but in your comment to Math Lover' s answer ending with "In that case, $a$ and $b$ must be integers" you said "$a,b$ can be real numbers". This confused me. Perhaps you edit once more and write "4 consecutive integers".
$endgroup$
– Paul Frost
Dec 4 '18 at 18:35












$begingroup$
The problem / task says: $a, b in Bbb R$, ... the roots of the equations are 4 consecutive numbers. I guess what Math Lover was saying is that the roots won't be integers if $a, b$ are not integers.
$endgroup$
– Pero
Dec 4 '18 at 18:38




$begingroup$
The problem / task says: $a, b in Bbb R$, ... the roots of the equations are 4 consecutive numbers. I guess what Math Lover was saying is that the roots won't be integers if $a, b$ are not integers.
$endgroup$
– Pero
Dec 4 '18 at 18:38










3 Answers
3






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1












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You made a mistake while solving $b$. Specifically, $b=1$. As such, $$a^2-4a-5 = (a-5)(a+1)=0 implies a = -1,5.$$



Regarding $x_2 < x_4 < x_3 < x_1$, you could argue by using the following argument.



Note that one of the roots is the minimum among them. Say that is $x_2$. Since
$$x_1 + x_2 = a = x_3 + x_4,$$
we have
$$x_1 = x_3 + (x_4 - x_2) implies x_1 > x_3$$
because $x_4 > x_2$. Likewise, $x_1 > x_4$.





You are looking for integer solutions. In that case, $a$ and $b$ must be integers






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$endgroup$













  • $begingroup$
    $a, b$ can be real numbers.
    $endgroup$
    – Pero
    Dec 4 '18 at 18:05










  • $begingroup$
    Oh wait $a, b in Bbb R$, but because they are consecutive they must be integers right?
    $endgroup$
    – Pero
    Dec 4 '18 at 18:28










  • $begingroup$
    The $x_i$ are integers, hence by Vieta's formulas $a, b$ must be integers.
    $endgroup$
    – Paul Frost
    Dec 4 '18 at 18:39



















1












$begingroup$

Let four integer roots be $n-1,n,n+1,n+2$. Write
$$ (x^2+ax+a-b)(x^2+ax+a+b)=(x-n+1)(x-n)(x-n-1)(x-n-2).$$
or
$$ x^4+2ax^3+(a^2+2a)x^2+2a^2x+(a^2-b^2)=x^4-2(2n+1)x^3+(6n^2+6n-1)x^2-2(2n^3+3n^2-n-1)x+(n^4+2n^3-n^2-2n). tag{1} $$
Comparing the coefficients of $x^3$ and $x^2$ and the constants of (1) gives
$$ 2a=-2(2n+1), a^2+2a=6n^2+6n-1, a^2-b^2=n^4+2n^3-n^2-2n.$$
From these equations, one can solve
$$ a=-1,b=pm1,n=0,text{ or } a=5,,b=pm1,n=-3.$$






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    $begingroup$

    As pointed out by Math Lover, you made a mistake in your computations. The correct solutions for $a, b$ are $a = -1,5$ and $b = 1$. But if you look at xpaul's answer, there are two more solutions $a = -1,5$ and $b = -1$. Where do they come from?



    Observe that neccessarily $b ne 0$, otherwise we would not obtain 4 solutions. But if we get 4 solutions for some $b$, we get the same solutions if we replace $b$ by $-b$ (this replacement is equivalent to permuting the two equations). That explains the occurrence of $a = -1,5$ and $b = -1$. But why didn't you find them?



    This comes from your ordering assumption $x_2 < x_4 < x_1 < x_3$. We have $x_{1/2} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a + 4b}$, $x_{3/4} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a - 4b}$. Now let us assume $b > 0$. Then we see that $x_2 < x_4 < -frac{1}{2}a < x_3 < x_1$. It also shows that your ordering is not correct, although it does not matter for your further computations. If we assume $b < 0$, we get $x_4 < x_2 < x_1 < x_3$ which leads to the "missing" solutions for $a, b$.



    Finally note that $a = -1$ and $b = pm1$ yield the 4 solutions $-1,0,1,2$ and $a = 5$ and $b = pm1$ yield $-4,-3,-2,-1$.






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      3 Answers
      3






      active

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      You made a mistake while solving $b$. Specifically, $b=1$. As such, $$a^2-4a-5 = (a-5)(a+1)=0 implies a = -1,5.$$



      Regarding $x_2 < x_4 < x_3 < x_1$, you could argue by using the following argument.



      Note that one of the roots is the minimum among them. Say that is $x_2$. Since
      $$x_1 + x_2 = a = x_3 + x_4,$$
      we have
      $$x_1 = x_3 + (x_4 - x_2) implies x_1 > x_3$$
      because $x_4 > x_2$. Likewise, $x_1 > x_4$.





      You are looking for integer solutions. In that case, $a$ and $b$ must be integers






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        $a, b$ can be real numbers.
        $endgroup$
        – Pero
        Dec 4 '18 at 18:05










      • $begingroup$
        Oh wait $a, b in Bbb R$, but because they are consecutive they must be integers right?
        $endgroup$
        – Pero
        Dec 4 '18 at 18:28










      • $begingroup$
        The $x_i$ are integers, hence by Vieta's formulas $a, b$ must be integers.
        $endgroup$
        – Paul Frost
        Dec 4 '18 at 18:39
















      1












      $begingroup$

      You made a mistake while solving $b$. Specifically, $b=1$. As such, $$a^2-4a-5 = (a-5)(a+1)=0 implies a = -1,5.$$



      Regarding $x_2 < x_4 < x_3 < x_1$, you could argue by using the following argument.



      Note that one of the roots is the minimum among them. Say that is $x_2$. Since
      $$x_1 + x_2 = a = x_3 + x_4,$$
      we have
      $$x_1 = x_3 + (x_4 - x_2) implies x_1 > x_3$$
      because $x_4 > x_2$. Likewise, $x_1 > x_4$.





      You are looking for integer solutions. In that case, $a$ and $b$ must be integers






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        $a, b$ can be real numbers.
        $endgroup$
        – Pero
        Dec 4 '18 at 18:05










      • $begingroup$
        Oh wait $a, b in Bbb R$, but because they are consecutive they must be integers right?
        $endgroup$
        – Pero
        Dec 4 '18 at 18:28










      • $begingroup$
        The $x_i$ are integers, hence by Vieta's formulas $a, b$ must be integers.
        $endgroup$
        – Paul Frost
        Dec 4 '18 at 18:39














      1












      1








      1





      $begingroup$

      You made a mistake while solving $b$. Specifically, $b=1$. As such, $$a^2-4a-5 = (a-5)(a+1)=0 implies a = -1,5.$$



      Regarding $x_2 < x_4 < x_3 < x_1$, you could argue by using the following argument.



      Note that one of the roots is the minimum among them. Say that is $x_2$. Since
      $$x_1 + x_2 = a = x_3 + x_4,$$
      we have
      $$x_1 = x_3 + (x_4 - x_2) implies x_1 > x_3$$
      because $x_4 > x_2$. Likewise, $x_1 > x_4$.





      You are looking for integer solutions. In that case, $a$ and $b$ must be integers






      share|cite|improve this answer











      $endgroup$



      You made a mistake while solving $b$. Specifically, $b=1$. As such, $$a^2-4a-5 = (a-5)(a+1)=0 implies a = -1,5.$$



      Regarding $x_2 < x_4 < x_3 < x_1$, you could argue by using the following argument.



      Note that one of the roots is the minimum among them. Say that is $x_2$. Since
      $$x_1 + x_2 = a = x_3 + x_4,$$
      we have
      $$x_1 = x_3 + (x_4 - x_2) implies x_1 > x_3$$
      because $x_4 > x_2$. Likewise, $x_1 > x_4$.





      You are looking for integer solutions. In that case, $a$ and $b$ must be integers







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 4 '18 at 17:51

























      answered Dec 4 '18 at 17:35









      Math LoverMath Lover

      14k31436




      14k31436












      • $begingroup$
        $a, b$ can be real numbers.
        $endgroup$
        – Pero
        Dec 4 '18 at 18:05










      • $begingroup$
        Oh wait $a, b in Bbb R$, but because they are consecutive they must be integers right?
        $endgroup$
        – Pero
        Dec 4 '18 at 18:28










      • $begingroup$
        The $x_i$ are integers, hence by Vieta's formulas $a, b$ must be integers.
        $endgroup$
        – Paul Frost
        Dec 4 '18 at 18:39


















      • $begingroup$
        $a, b$ can be real numbers.
        $endgroup$
        – Pero
        Dec 4 '18 at 18:05










      • $begingroup$
        Oh wait $a, b in Bbb R$, but because they are consecutive they must be integers right?
        $endgroup$
        – Pero
        Dec 4 '18 at 18:28










      • $begingroup$
        The $x_i$ are integers, hence by Vieta's formulas $a, b$ must be integers.
        $endgroup$
        – Paul Frost
        Dec 4 '18 at 18:39
















      $begingroup$
      $a, b$ can be real numbers.
      $endgroup$
      – Pero
      Dec 4 '18 at 18:05




      $begingroup$
      $a, b$ can be real numbers.
      $endgroup$
      – Pero
      Dec 4 '18 at 18:05












      $begingroup$
      Oh wait $a, b in Bbb R$, but because they are consecutive they must be integers right?
      $endgroup$
      – Pero
      Dec 4 '18 at 18:28




      $begingroup$
      Oh wait $a, b in Bbb R$, but because they are consecutive they must be integers right?
      $endgroup$
      – Pero
      Dec 4 '18 at 18:28












      $begingroup$
      The $x_i$ are integers, hence by Vieta's formulas $a, b$ must be integers.
      $endgroup$
      – Paul Frost
      Dec 4 '18 at 18:39




      $begingroup$
      The $x_i$ are integers, hence by Vieta's formulas $a, b$ must be integers.
      $endgroup$
      – Paul Frost
      Dec 4 '18 at 18:39











      1












      $begingroup$

      Let four integer roots be $n-1,n,n+1,n+2$. Write
      $$ (x^2+ax+a-b)(x^2+ax+a+b)=(x-n+1)(x-n)(x-n-1)(x-n-2).$$
      or
      $$ x^4+2ax^3+(a^2+2a)x^2+2a^2x+(a^2-b^2)=x^4-2(2n+1)x^3+(6n^2+6n-1)x^2-2(2n^3+3n^2-n-1)x+(n^4+2n^3-n^2-2n). tag{1} $$
      Comparing the coefficients of $x^3$ and $x^2$ and the constants of (1) gives
      $$ 2a=-2(2n+1), a^2+2a=6n^2+6n-1, a^2-b^2=n^4+2n^3-n^2-2n.$$
      From these equations, one can solve
      $$ a=-1,b=pm1,n=0,text{ or } a=5,,b=pm1,n=-3.$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let four integer roots be $n-1,n,n+1,n+2$. Write
        $$ (x^2+ax+a-b)(x^2+ax+a+b)=(x-n+1)(x-n)(x-n-1)(x-n-2).$$
        or
        $$ x^4+2ax^3+(a^2+2a)x^2+2a^2x+(a^2-b^2)=x^4-2(2n+1)x^3+(6n^2+6n-1)x^2-2(2n^3+3n^2-n-1)x+(n^4+2n^3-n^2-2n). tag{1} $$
        Comparing the coefficients of $x^3$ and $x^2$ and the constants of (1) gives
        $$ 2a=-2(2n+1), a^2+2a=6n^2+6n-1, a^2-b^2=n^4+2n^3-n^2-2n.$$
        From these equations, one can solve
        $$ a=-1,b=pm1,n=0,text{ or } a=5,,b=pm1,n=-3.$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let four integer roots be $n-1,n,n+1,n+2$. Write
          $$ (x^2+ax+a-b)(x^2+ax+a+b)=(x-n+1)(x-n)(x-n-1)(x-n-2).$$
          or
          $$ x^4+2ax^3+(a^2+2a)x^2+2a^2x+(a^2-b^2)=x^4-2(2n+1)x^3+(6n^2+6n-1)x^2-2(2n^3+3n^2-n-1)x+(n^4+2n^3-n^2-2n). tag{1} $$
          Comparing the coefficients of $x^3$ and $x^2$ and the constants of (1) gives
          $$ 2a=-2(2n+1), a^2+2a=6n^2+6n-1, a^2-b^2=n^4+2n^3-n^2-2n.$$
          From these equations, one can solve
          $$ a=-1,b=pm1,n=0,text{ or } a=5,,b=pm1,n=-3.$$






          share|cite|improve this answer









          $endgroup$



          Let four integer roots be $n-1,n,n+1,n+2$. Write
          $$ (x^2+ax+a-b)(x^2+ax+a+b)=(x-n+1)(x-n)(x-n-1)(x-n-2).$$
          or
          $$ x^4+2ax^3+(a^2+2a)x^2+2a^2x+(a^2-b^2)=x^4-2(2n+1)x^3+(6n^2+6n-1)x^2-2(2n^3+3n^2-n-1)x+(n^4+2n^3-n^2-2n). tag{1} $$
          Comparing the coefficients of $x^3$ and $x^2$ and the constants of (1) gives
          $$ 2a=-2(2n+1), a^2+2a=6n^2+6n-1, a^2-b^2=n^4+2n^3-n^2-2n.$$
          From these equations, one can solve
          $$ a=-1,b=pm1,n=0,text{ or } a=5,,b=pm1,n=-3.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 18:05









          xpaulxpaul

          22.7k24455




          22.7k24455























              1












              $begingroup$

              As pointed out by Math Lover, you made a mistake in your computations. The correct solutions for $a, b$ are $a = -1,5$ and $b = 1$. But if you look at xpaul's answer, there are two more solutions $a = -1,5$ and $b = -1$. Where do they come from?



              Observe that neccessarily $b ne 0$, otherwise we would not obtain 4 solutions. But if we get 4 solutions for some $b$, we get the same solutions if we replace $b$ by $-b$ (this replacement is equivalent to permuting the two equations). That explains the occurrence of $a = -1,5$ and $b = -1$. But why didn't you find them?



              This comes from your ordering assumption $x_2 < x_4 < x_1 < x_3$. We have $x_{1/2} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a + 4b}$, $x_{3/4} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a - 4b}$. Now let us assume $b > 0$. Then we see that $x_2 < x_4 < -frac{1}{2}a < x_3 < x_1$. It also shows that your ordering is not correct, although it does not matter for your further computations. If we assume $b < 0$, we get $x_4 < x_2 < x_1 < x_3$ which leads to the "missing" solutions for $a, b$.



              Finally note that $a = -1$ and $b = pm1$ yield the 4 solutions $-1,0,1,2$ and $a = 5$ and $b = pm1$ yield $-4,-3,-2,-1$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                As pointed out by Math Lover, you made a mistake in your computations. The correct solutions for $a, b$ are $a = -1,5$ and $b = 1$. But if you look at xpaul's answer, there are two more solutions $a = -1,5$ and $b = -1$. Where do they come from?



                Observe that neccessarily $b ne 0$, otherwise we would not obtain 4 solutions. But if we get 4 solutions for some $b$, we get the same solutions if we replace $b$ by $-b$ (this replacement is equivalent to permuting the two equations). That explains the occurrence of $a = -1,5$ and $b = -1$. But why didn't you find them?



                This comes from your ordering assumption $x_2 < x_4 < x_1 < x_3$. We have $x_{1/2} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a + 4b}$, $x_{3/4} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a - 4b}$. Now let us assume $b > 0$. Then we see that $x_2 < x_4 < -frac{1}{2}a < x_3 < x_1$. It also shows that your ordering is not correct, although it does not matter for your further computations. If we assume $b < 0$, we get $x_4 < x_2 < x_1 < x_3$ which leads to the "missing" solutions for $a, b$.



                Finally note that $a = -1$ and $b = pm1$ yield the 4 solutions $-1,0,1,2$ and $a = 5$ and $b = pm1$ yield $-4,-3,-2,-1$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  As pointed out by Math Lover, you made a mistake in your computations. The correct solutions for $a, b$ are $a = -1,5$ and $b = 1$. But if you look at xpaul's answer, there are two more solutions $a = -1,5$ and $b = -1$. Where do they come from?



                  Observe that neccessarily $b ne 0$, otherwise we would not obtain 4 solutions. But if we get 4 solutions for some $b$, we get the same solutions if we replace $b$ by $-b$ (this replacement is equivalent to permuting the two equations). That explains the occurrence of $a = -1,5$ and $b = -1$. But why didn't you find them?



                  This comes from your ordering assumption $x_2 < x_4 < x_1 < x_3$. We have $x_{1/2} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a + 4b}$, $x_{3/4} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a - 4b}$. Now let us assume $b > 0$. Then we see that $x_2 < x_4 < -frac{1}{2}a < x_3 < x_1$. It also shows that your ordering is not correct, although it does not matter for your further computations. If we assume $b < 0$, we get $x_4 < x_2 < x_1 < x_3$ which leads to the "missing" solutions for $a, b$.



                  Finally note that $a = -1$ and $b = pm1$ yield the 4 solutions $-1,0,1,2$ and $a = 5$ and $b = pm1$ yield $-4,-3,-2,-1$.






                  share|cite|improve this answer









                  $endgroup$



                  As pointed out by Math Lover, you made a mistake in your computations. The correct solutions for $a, b$ are $a = -1,5$ and $b = 1$. But if you look at xpaul's answer, there are two more solutions $a = -1,5$ and $b = -1$. Where do they come from?



                  Observe that neccessarily $b ne 0$, otherwise we would not obtain 4 solutions. But if we get 4 solutions for some $b$, we get the same solutions if we replace $b$ by $-b$ (this replacement is equivalent to permuting the two equations). That explains the occurrence of $a = -1,5$ and $b = -1$. But why didn't you find them?



                  This comes from your ordering assumption $x_2 < x_4 < x_1 < x_3$. We have $x_{1/2} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a + 4b}$, $x_{3/4} = -frac{1}{2}a pm frac{1}{2}sqrt{a^2 - 4a - 4b}$. Now let us assume $b > 0$. Then we see that $x_2 < x_4 < -frac{1}{2}a < x_3 < x_1$. It also shows that your ordering is not correct, although it does not matter for your further computations. If we assume $b < 0$, we get $x_4 < x_2 < x_1 < x_3$ which leads to the "missing" solutions for $a, b$.



                  Finally note that $a = -1$ and $b = pm1$ yield the 4 solutions $-1,0,1,2$ and $a = 5$ and $b = pm1$ yield $-4,-3,-2,-1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 20:03









                  Paul FrostPaul Frost

                  10.3k3933




                  10.3k3933






























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