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Find the sum of the values of the parameter $a$ such that for some non-constant polynomials with real coefficients $f(x)$ and $g(x)$,

$$g(x)^2 =f(x)^3−12 f(x)^2 + 45 f(x) − a$$

My understanding: Since $g(x)^2$ is always positive, $$f(x)^3−12f(x)^2+45f(x)−a \geq0.$$

$f(x)^3−12f(x)^2+45f(x) \geq 0$ for $x \geq 0$ and $<0$ for $x<0$.Therefore, for $x\geq0$, $a \in (-\infty,0]$ and for $x<0$, $a > 0$.

Hence the sum should be zero, since $a$ can take all the possible values. Can anyone please tell me whether the above is correct or not?

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If $a$ is fixed at the beginning, you can't let it move afterwards. Unless I did not understand the question. –  1015 Mar 6 '13 at 18:30
    
What values can $a$ take? Real, integer, ...? –  vonbrand Mar 6 '13 at 20:49
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it can take any value –  user65368 Mar 6 '13 at 21:01
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This is a problem posed by Brilliant, which provides weekly problems for students to work on. You may view it here - Solve this problem: Poly-polynomials. As the week is over, I'm fine with the posts being kept up. It does explain why there are repeat versions of this problem, and it will unlikely get accepted. - Calvin Lin Brilliant Challenge Master –  Calvin Lin Mar 11 '13 at 20:08

3 Answers 3

Lemma For $\alpha_1, \alpha_2, \ldots, \alpha_n \in \mathbb{C}$ such that $\alpha_1 \ne \alpha_2 \ne \cdots \ne \alpha_n$ and $p(z), q(z)$ any two polynomials over $\mathbb{C}$: $$p(z)^2 = \prod_{i=1}^n(q(z)-\alpha_i) \implies p(z), q(z) \text{ constant }\tag{*}$$

(Proof) Assume the contrary, suppose there are two non-constant polynomials $p(z), q(z)$ such that $$p(z)^2 = \prod_{i=1}^n(q(z)-\alpha_i)$$ Since $\alpha_1, \alpha_2, \ldots, \alpha_n$ are all distinct, the set of roots for $q(z) - \alpha_i, i = 1,\ldots,n$ are disjoint from each other. Since the roots on L.H.S of $(*)$ appear in pairs and $\mathbb{C}[x]$ is an UFD. The roots of each $q(z) - \alpha_i$ also come in pairs. This means we can factor $p(z)$ into $n$ pieces such that:

$$p(z) = \prod_{i=1}^n p_i(z) \;\;\text{ and }\;\; p_i(z)^2 = q(z) - \alpha_i,\, i = 1,\ldots,n$$

Notice for $1 \le i \ne j \le n$, $$(p_i(z) + p_j(z))(p_i(z) - p_j(z)) = (q(z) - \alpha_i) - ( q(z) - \alpha_j ) = \alpha_j - \alpha_i \ne 0$$ are constants. This implies all $p_i(z)$ are constants. As a result, $p(z) = \prod_{i=1}^n p_i(z)$ and $q(z) = p_1(z)^2 + \alpha_1$ are also constants, a contradiction!

(Comment) In the general case when the multiplicity of some $\alpha_i$ is greater than $1$. A similar argument shows that in order for $$ p(z)^2 = \prod_{i=1}^n (q(z) - \alpha_i)^{m_i}$$ to have non-constant solutions, a necessary and sufficient condition is at most one of the multiplicity $m_i$ is odd. However, we don't need to use this piece of info for the original problem.

Back to original problem

Applying the lemma to the equation $$g(x)^2 = P(f(x)) = f(x)^3−12 f(x)^2 + 45 f(x) − a\tag{**}$$ In order for it to has non-constant solutions, a necessary condition is the cubic polynomial $P(z) = z^3 - 12z^2 + 45z - a$ has repeated roots. Let $c$ be any repeated root, $(z - c)^2$ will be a factor of $P(z)$ and hence $P'(c) = 3c^2 - 24 c + 45 = 3(c-3)(c-5) = 0 \implies c = 3 \text{ or } 5$.

Case c = 3

$P(3) = 0 \implies a = 54$ and $(**)$ reduces to $g^2 = P(f) = (f-6)(f-3)^2$.

If one assign $6,3,3$ to $\alpha_1,\alpha_2,\alpha_3$ and repeat the arguments in proof of lemma, we see $g(z)$ contains a factor $h(z)$ such that:

$$h(z)^2 = f(z) - 6 \implies g(z)^2 = P(f(z)) = h(z)^2(h(z)^2 + 6 - 3)^2$$

This implies $(**)$ has solutions of the form

$$\begin{align} f(z) &= h(z)^2 + 6\\ g(z) &= h(z)(h(z)^2 + 3) \end{align}$$

Case c = 5

$P(5) = 0 \implies a = 50$ and $(**)$ reduces to $g^2 = P(f) = (f-2)(f-5)^2$.

Once again $g(z)$ contains a factor $h(z)$ such that:

$$h(z)^2 = f(z) - 2 \implies g(z)^2 = P(f(z)) = h(z)^2(h(z)^2 + 2 - 5)^2$$

This implies $(**)$ has solutions of the form

$$\begin{align} f(z) &= h(z)^2 + 2\\ g(z) &= h(z)(h(z)^2 - 3) \end{align}$$

And the desired answer is $54 + 50 = 104$.

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And there's still no upvote...+1. –  1015 Mar 7 '13 at 20:57

This makes no sense: there's no reason that the sign of $f(x)^3 - 12 f(x)^2 + 45 f(x)$ should have anything to do with the sign of $x$.

In any case, you want $f(x)^3 - 12 f(x)^2 + 45 f(x) - a \ge 0$ for all $x$. That can only happen if $f$ is a polynomial of even degree.

As a matter of fact, if you want $f$ to be a quadratic polynomial (and thus $g$ cubic), you need $a = 50$ or $54$:

$$ \left( {x}^{2}+2 \right) ^{3}-12\, \left( {x}^{2}+2 \right) ^{2}+45\, \left({x}^{2}+2\right) - 50= \left( {x}^{3}-3\,x \right) ^{2} $$

$$\left( {x}^{2}+6 \right) ^{3}-12\, \left( {x}^{2}+6 \right) ^{2}+45\, \left({x}^{2}+6 \right) - 54= \left( {x}^{3}+3\,x \right) ^{2} $$

I don't know if there are solutions of higher degree.

EDIT: Let $P(z) = z^3 - 12 z^2 + 45 z - a$. Suppose $P(f(x)) = g(x)^2$, where $f$ and $g$ are nonconstant polynomials. If $g$ has degree $d$, then $f$ has degree $2d/3 < d$.

Differentiating the equation $P(f(x)) = g(x)^2$, we get $P'(f(x)) f'(x) = 2 g(x) g'(x)$. In particular, if $r$ is a (real or complex) root of $g(x)$, then either $P'(f(r)) = 0$ or $f'(r) = 0$. If $P'(f(r)) = 0$, then since the roots of $P'(z) = 3 z^2 - 24 z + 45 = 3 (z-3)(z-5)$ are $3$ and $5$, we have $f(r) = 3$ or $f(r) = 5$. Thus $P(3) = 0$ or $P(5) = 0$. But $P(3) = 54 - a$ and $P(5) = 50 - a$, so $a = 50$ or $54$.

Now it is possible that $f'(r) = 0$ for all roots $r$ of $g(x)$. But $f'$ has degree $< d - 1$, so there is at least one root $r$ which has higher multiplicity as a root of $g$ than it does as a root for $f'$. That is, for some $r$, $c \ne 0$ and positive integer $k$ we have $f(x) = f(r) + c (x-r)^{k} + O((x-r)^{k+1})$ while $g(x) = O((x-r)^{k})$. Then $$P(f(x)) = P'(f(r)) c (x-r)^k + O((x-r)^{k+1}) = g(x)^2 = O((x-r)^{k+1})$$ so again $P'(f(r)) = 0$.

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In fact, when $a=50, 54$, there are solutions of higher degree, but there are no solutions for all other $a$. I'll be posting an answer regarding this when I get home in a while. –  Ivan Loh Mar 7 '13 at 8:01

It is easy to plot $P(x)=x^3-12x^2+45x$. (This can also be done manually, by looking at critical points) Looking at the behaviour of $P(x)$, we note that the cubic equation $P(x)=a$ has 1 real root and 2 distinct complex roots (conjugate pair) for $a<50, a>54$, 3 distinct real roots for $50<a<54$, and 2 real roots (with 1 repeated) for $a=50, 54$.

Consider $a \not =50, 54$. Then the cubic equation $P(x)=a$ has at least 1 real root $c$ and no repeated complex roots. Factorise $g(x)^2=P(f(x))-a=(f(x)-c)(f(x)^2+pf(x)+q)$, where $p, q \in \mathbb{R}$.

We show that $(f(x)-c)$ and $(f(x)^2+pf(x)+q)$ have no common complex root. Assume on the contrary that they have a common root $\alpha$, then $f(\alpha)$ is a common complex root of $x-c$ and $x^2+px+q$, so $f(\alpha)$ is a repeated complex root of $P(x)=a$, a contradiction.

Now each complex root in $g(x)^2$ has even multiplicity, so each complex root in $(f(x)^2+pf(x)+q)$ has even multiplicity. Since $(f(x)^2+pf(x)+q) \in \mathbb{R}[x]$ and $(f(x)^2+pf(x)+q)$ has positive leading coefficient, this implies that $(f(x)^2+pf(x)+q)=h(x)^2$ for some $h(x) \in \mathbb{R}[x]$. (To see why, note that the complex roots come in conjugate pairs, so that $(f(x)^2+pf(x)+q)$ is the product of squares of quadratics.)

Finally, observe that since $x^2+px+q$ has no repeated roots, we have $x^2+px+q=(x+\frac{p}{2})^2+r$, where $r \not =0, r \in \mathbb{R}$. Thus $h(x)^2=(f(x)^2+pf(x)+q)=(f(x)+\frac{p}{2})^2+r$, so $r=(h(x)-(f(x)+\frac{p}{2}))(h(x)+(f(x)+\frac{p}{2}))$, so both $(h(x)-(f(x)+\frac{p}{2}))$ and $(h(x)+(f(x)+\frac{p}{2}))$ are constant, so that $f(x)$ must be constant, a contradiction.

For $a=50$, factorise $g(x)^2=P(f(x))-50=(f(x)-2)(f(x)-5)^2$, so we have the general family of solutions $f(x)=q(x)^2+2, g(x)=q(x)(f(x)-5)$, where we restrict $q(x)$ to be nonconstant, since both $f(x), g(x)$ are nonconstant.

For $a=54$, factorise $g(x)^2=P(f(x))-54=(f(x)-6)(f(x)-3)^2$, so we have the general family of solutions $f(x)=q(x)^2+6, g(x)=q(x)(f(x)-3)$, where we restrict $q(x)$ to be nonconstant, since both $f(x), g(x)$ are nonconstant.

Edit: Forgot to answer the question. To conclude, the sum of values of permissible $a$ is $50+54=104$.

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