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I need some help with the following problem. Many thanks in advance.

Let $f(x) = x^2+px+q$ and $g(x) = x^2+rx+s$. Find an expression for $f(g(x))$ and hence find a necessary and sufficient condition on $a$, $b$, $c$ for it to be possible to write the quartic expression $x^4+ax^3+bx^2+cx+d$ in the form $f(g(x))$, for some choice of values of $p$, $q$, $r$, and $s$.

Okay the first thing I did was to find $f(g(x))$:

$\begin{aligned}f(g(x)) & = (x^2+rx+s)^2+p(x^2+rx+s)+q \\& = x^2(x^2+rx+s)^2+rx(x^2+rx+s)+s(x^2+rx+s)+px^2+prx+ps+q \\& = x^4+rx^3+sx^2+rx^3+r^2+x^2+rsx+sx^2+srx+s^2+px^2+prx+ps+q \\& = x^4+(2r)x^3+(p+2s+r^2)x^2+(2rs+pr)x+s^2+q+ps \end{aligned}$

So we wish to have:

$x^4+(2r)x^3+({p}+2s+r^2)x^2+(2rs+pr)x+s^2+q+ps \equiv x^4+ax^3+bx^2+cx+d$

Comparing the coefficients, $r = \frac{1}{2}a$, $2s = b-\frac{1}{4}a^{2}-{p}$, and $c = r(2s+p)$, thus $c = \frac{ab}{2}-\frac{1}{{8}}a^{{3}} $.

I understand that this condition is 'necessary' -- my problem is that I'm not quite sure how to make it sufficient. I'm not quite sure how I'm supposed to choose some suitable values of p, q, r and s.

Show further that this condition holds if and only if it's possible to write the quartic expression $x^4+ax^3+bx^2+cx+d$ in the form $(x^2+vx+w)^2-k$, for some choice of values v, w, q, r, s.

$\begin{aligned} (x^2+vx+w)^2-k & = x^2(x^2+vx+w)+vx(x^2+vx+w)+w(x^2+vx+w)-k \\& = x^4+vx^3+wx^2+vx^3+vx^2+wvx+wx^2+vwx+w^2-k \\& = x^4+(2v)x^3+(2w+v^2)x^2+(2vw)x+w^2-k. \end{aligned}$

I see that the 'suitable choice' would have been q = 0, but how was I supposed to see that?

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    I don't think your ccondition on $c$ follows. $\frac{ab}{2} - \frac{1}{8}a^2$ is $2rs$, so you seem to be saying that $rp = -\frac{1}{8}a^2$, hence that $p=-\frac{1}{4}a$; on what grounds? Also, you never addressed $d$ or $q$.2011-06-17
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    To prove your condition sufficient, you should be able to take any $a,b,c,d$ satisfying your condition(s) and derive $p,q,r,s$2011-06-18
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    @Arturo Magidin. My apology! I've made a mistake when typing and omitted a p. I highlighted the change with blue fonts. As for considering d or q, the question asked me to find the condition on a, b, c. It then wants me to find a suitable value in p, q, r and s such that the condition is sufficient. Or at least that's how I understood it.2011-06-18
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    @Ross Millikan Thank you! Let me try that...2011-06-18
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    Shouldn't the last term in your expression for $c$ be $-\frac{1}{8}a^3$?2011-06-18
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    You're right, fixed that too!2011-06-18
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    @Ross Millikan, Okay, I think I get the problem now. So if I take $a = 2, b = 3, c = 2$ -- then I get $r = 1$. Also $2 = p+2s$ and $2 = p+2s$. Writing $s = 1-p/2$, we take any value of $p$, say, $p = 4$; then we have $s = -1$ and since $s²+q+ps = 0$, we have $q = 3$.2011-06-18
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    So given $a,b,d$, you can calculate $p+2s,r,$ and $s^2+q+ps$. You have a degree of freedom left. As you say, you can pick $p$ and then calculate $q,s$. For the second part, you can just show that the coefficients in your expansion satisfy the same restriction on $a,b,c$. In the last line, it should be $k$, not $k^2$2011-06-18
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    @Ross Millikan - Thank you. So for the second part, the following suffices... $$\begin{aligned} & \frac{(2v)(2w+v^2)}{2}-\frac{1}{8}(2v)^3 \\& = \frac{8v(2w+v^2)-8v^3}{8} \\& = \frac{16vw+8v^3-8v^3}{8} \\& = \frac{16vw}{8} \\& = 2vw.\end{aligned}$$ ... as it satisfies our necessary and sufficient condition, and hence the 'if and if only if'.2011-06-18
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    Sorry about the typos/mistakes in the main body of this problem. I've an exam in few days so I sort of panicked when I couldn't do this one.2011-06-18

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