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Let $f(x) = x^3 + a x^2 + b x + c$ and $g(x) = x^3 + b x^2 + c x + a\,$ where $a, b, c$ are integers and $c\neq 0\,$. Suppose that the following conditions hold:

  1. $f(1)=0$
  2. The roots of $g(x)$ are squares of the roots of $f(x)$.

I'd like to find $a, b$ and $c$.

I tried solving equations made using condition 1. and relation between the roots, but couldn't solve. The equation which I got in $c$ is $c^4 + c^2 +3 c-1=0$ (edit: eqn is wrong). Also I was able to express $a$ and $b$ in terms of $c$. But the equation isn't solvable by hand.

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  • $\begingroup$ Welcome to MSE. Click on the "edit" button below your question to add details, like exactly what equation you derived from conditions 1 and 2, and how you attempted to solve them; then you'll be more likely to get help that's well-targeted. Simple "do my homework for me questions" don't get nearly as much traction here as ones that show you've done some work yourself (and therefore don't waste our time giving detailed answers when it's a simple matter of a missing minus-sign, for instance). $\endgroup$ – John Hughes Apr 3 at 11:32
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    $\begingroup$ Which equations could you write down? $\endgroup$ – Math-fun Apr 3 at 11:35
  • $\begingroup$ Also: I've edited to clean up the MathJax: in general, go ahead and put anything math-like between dollar signs. Rather than things like $a$ = $b$x + $c$ which produces $a$ = $b$x + $c$ let MathJax do its magic at formatting equations: $a = bx + c $ will produce $a = bx + c$. $\endgroup$ – John Hughes Apr 3 at 11:35
  • $\begingroup$ Edited with Eq. $\endgroup$ – Meet Shah Apr 3 at 11:36
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    $\begingroup$ I don't understand how you got that equation in $c$. And if you were "able to express $a$ and $b$ in terms of $c$," why not tell us what the expression was? Go ahead and type a little more, so that we understand what you know/don't know and what you can and cannot do. $\endgroup$ – John Hughes Apr 3 at 11:44
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Let $u$, $v$ and $w$ be the roots of $f$, so that $u^2$, $v^2$ and $w^2$ are the roots of $g$. Then comparing the coefficients of $$(x-u)(x-v)(x-w)=f(x)=x^3+ax^2+bx+c,$$ $$(x-u^2)(x-v^2)(x-w^2)=g(x)=x^3+bx^2+cx+a,$$ yields the equations \begin{eqnarray*} a&=&-u-v-w&=&-u^2v^2w^2,\\ b&=&uv+uw+vw&=&-u^2-v^2-w^2,\\ c&=&-uvw&=&u^2v^2+u^2w^2+v^2w^2. \end{eqnarray*} This immediately shows that $a=-c^2$, and the identities \begin{eqnarray*} u^2+v^2+w^2&=&(u+v+w)^2-2(uv+uw+vw),\\ u^2v^2+u^2w^2+v^2w^2&=&uvw(u+v+w)-(uv+uw+vw)^2, \end{eqnarray*} show that $-b=a^2-2b$ and $c=ac-b^2$, respectively, hence $b=a^2=c^4$ and so $$f(x)=x^3-c^2x^2+c^4x+c,$$ for some $c$. Then $f(1)=1$ implies that $$c^4-c^2+c+1=0,$$ which has the clear root $c=-1$. Then $a=-1$ and $b=1$.

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Hint: Condition 2 can be expressed as $f(x)$ divides $g(x^2)$.

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The conditions 1 and 2 imply that

  • $\;c\,$ must be $\,0\,$ or $\,-1$,
  • $\;a= -c^2$, and $\:b= c^2-c-1\,$.

Condition 1 gives $0=f(1)=1+a+b+c=g(1)\,.\,$ Hence both $f$ and $g$ have a zero at $1$ and factor as $$f(x) \,=\, (x-1)\big(x^2 + (a+1)x -c\big)\\[1.5ex] g(x) \,=\, (x-1)\big(x^2 -(a+c)x -a\big)\,.$$ Denote the roots of the quadratic factor of $f$ by $x_1$ and $x_2$. Condition 2 says that the roots of $g$ are contained in $\{1,x_1^2,x_2^2\}$. By Vieta's formula one gets $$\,-a \,=\, x_1^2x_2^2=(-c)^2 \,=\, c^2\,,\;\text{thus}\;\; b \,=\, -a-c-1 \,=\, c^2-c-1\,.\tag{1}$$ Note that condition 2 remains true when restricted to the quadratic factors of $f$ and $\,g$. These are $$q_f \,=\,x^2 + \left(1-c^2\right)x -c\tag{2}\\ q_g \,=\,x^2 -c\,(1-c)x +c^2$$ when written in terms of $\,c$.
It is shown next that condition 2 cannot hold if $c\neq 0\,$ or $\,-1$.

  1. Assume $c\geqslant 1$. Then $q_f(0)=-c<0$, and the roots $x_1,x_2$ of $q_f$ are real and distinct. By Vieta's formula regarding $q_g$ one reaches the contradiction $0<x_1^2+x_2^2=c(1-c)\leq 0\,$.

  2. Assume $c\leq -2\,$. Then the discriminant $\left(1-c^2\right)^2+4c$ in $(2)$ is positive, and we run into the same contradiction as before.

So we are left with the two solutions (with condition 2 obviously satisfied)

  • $a=0,b=-1,c=0\:$ which was ruled out a priori
    then $f(x)=x(x+1)(x-1)\:$ and $\:g(x)=x^2(x-1)$
  • $a=-1,b=1,c=-1\:$
    where $f(x)=\left(x^2+1\right)(x-1)\:$ and $\:g(x)=(x+1)^2(x-1)$
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