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Find all $a, b \in \Bbb R$, ($b\ne0)$, 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: $$\frac{a^2-9}4=a-b,\quad \frac{a^2-1}4=a+b$$

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

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 $x_2\cdot x_4$ 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
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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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  • $\begingroup$ $a, b$ 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
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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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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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