Poly-polynomials

As with all polynomial problems, there are only two major ideas:

(i) Analysis of roots - whether it has double roots,etc. (a really good method would be to differentiate to bring down the highest power)

(ii) Analysis of degree

And not to forget, when doing algebra, we try our best to simplify the polynomial equations we are working with. In particular, the LHS of this equation looks very complicated, but notice that suppose we define a new polynomial $P(x) = x^3 - 12x^2 + 45x - a$ , then we can rewrite this equation neatly as:

$g(x)^2 = P(f(x)) (*)$

Now, we know nothing of the roots, so we apply idea (ii) and do some analysis on the degree. For instance, if $\deg g = d$ , then $\deg f = \frac{2d}{3} < d$ . We see no apparent use for this, so we leave this for later.

Next, we turn to idea (i), in the sleek form, we can differentiate $(*)$ to get:

$P'(f(x)) \times f'(x) = 2 g(x)g'(x)$

Aha, so if $q$ is a root of $g(x)$ , then either $P'(f(r)) = 0$ or $f'(r) = 0$ . We will attempt the former first, because it is considerably simpler.

So suppose $P'(f(r)) = 0$ . Firstly, we evaluate $P'(x)$ to get: $3x^2 - 24x + 45$ , being a quadratic equation, it is easy to see that it in fact can be simplified to the following: $3(x-3)(x-5)$ . Voila! We have that $f(r)$ is either $3$ or $5$ . In which case, we can simplify $P(3) = 0$ or $P(5) = 0$ . However, $P(3) = 54 - a$ and $P(5) = 50 -a$ so $a = \fbox{50,54}$ .

I basically was too lazy to work out the other part and tried $50 + 54 = \fbox{104}$ and it worked.

Now, with this "conjecture" in mind, we deal with the other case $f'(r) = 0$ . Notice that other than the fact of its degree, we know virtually nothing about the behaviour, etc. of $f(x)$ , which makes things tough. However, due to the "conjecture", we are motivated to try to reduce $f'(r) = 0$ to the above case of $P'(f(r)) = 0$ . So firstly, a list of things that we know:

(i) $\deg f < d-1$

And... that's all. However, we can compare this with $g$ . For instance, this clearly implies that there definitely exists (at least) a root $r$ that has higher multiplicity as a root of $g$ than as a root for $f′$ . One clever way to represent the growth rate/exponents, is to use Landau O-notation . So, we write (for some $n$ and positive $m$ ) $f(x) = f(r) + m(x-r)^n + O((x-r)^{n+1})$ and $g(x) = O((x-r)^{n})$ . Hmm, this looks better! So if we substitute everything back into $(*)$ , (remember that we want to reduce to $P'(f(r)) = 0$ ):

$P(f(x)) = P'(f(r))m(x-r)^n + O((x-r)^{n+1}) = O((x-r)^{n+1}) = g(x)^2$

So simplifying, we do indeed get that we need $P'(f(r)) = 0$ as desired.

Let's say $f(x)^3 - 12f(x)^2+45f(x) - a = (f(x)-c_1)(f(x)-c_2)(f(x)-c_3)$ Then divide into the following cases :

i) All the roots are the same ( $c_1 = c_2 =c_3)$ , from which we will get $3c_1 = 12 \Rightarrow c_1 = 4$ But then the right hand side will be $(f(x)-4)^3 = f(x)^3-12f(x)^2+48f(x)-64$ $\neq f(x)^3 - 12f(x)^2+45f(x) - a$ for any $a$ . Hence, no solution from this case.

ii) Exactly two of the roots are the same (WLOG, $c1 = c2)$ , from which we will get $2c_1+c_3 = 12$ $2c_1c_3+c_1^2 = 45$ From the 1st equation, substitute $c_3 = 12-2c_1$ to the 2nd equation, and we will get $3c_1^2-24c_1+45 = 0$ and thus the following pairs $(c_1,c_3) = (3,6),(5,2)$ . One can easily get $a = 54,50$ , respectively. And any of this $a$ will work since now $f(x)^3 - 12f(x)^2+45f(x) - a$ $= (f(x)-c_1)(f(x)-c_1)(f(x)-c_3)$ $= (f(x)-c_1)^2(f(x)-c_3)$ and we can choose $f(x) = h(x)^2+c_3$ for some polynomial $h$ so that the above's form is a square of a polynomial.

(iii) All the roots are different. In this case, let's say $p$ is a complex root of $g(x)$ , then we must have even power of $(x-p)$ in $g(x)^2$ and so in $(f(x)-c_1)(f(x)-c_2)(f(x)-c_3)$ too. But then, $p$ is the root of exactly one of $f(x)-c_1,f(x)-c_2,f(x)-c_3$ since $f(p)$ has exactly one value (taken from $c_1,c_2,c_3$ ) and therefore one of them has even power of $(x-p)$ as its factors. Moreover, this works for all roots of $g(x)$ , and so each of $f(x)-c_1,f(x)-c_2,f(x)-c_3$ is a square of polynomial, let's call it $f_1,f_2,f_3$ respectively. Now we have $f_1^2-f_2^2 = (f(x)-c_1)-(f(x)-c_2) = c_2-c_1 = c \neq 0$ $(f_1-f_2)(f_1+f_2) = c$ which forces $f_1,f_2$ to be constant functions, and so thus $f$ . Hence, no solution from this case, too. Therefore the only possible values of $a$ are $54,50$ and the sum is then $104$ .

Let N=x^3-12x^2+45x-a We will have N(f(x))=[g(x)]^2 If \deg g(x)=d then \deg f(x)=\frac{2d}{3} Differentating both side of the equation yields N'(f(x))f'(x)=2g'(x)g(x) If n is the root (real or complex) of g(x) then N'(f(n))=0 or f'(n)=0 Since \deg f(x)< \deg g(x) there is a number p such that g(p)=0 and f(p) \ne 0 So there must be one root n of g(x) which also the root of N(f'(x)) N'(x)=0 \Rightarrow x \in {3; 5} f(n) \in {3; 5} \Rightarrow a \in {50; 54} So our solution is 104

Let $T(x)=x^3-12x^2+45x-a$ then $T(f(x)) = [g(x)]^2\ \ \ (1)$ . Therefore if $g(x)$ has degree d, $f(x)$ has degree $\dfrac{2d}{3}$ .

Differentiating (1) we also get $$T'(f(x))f'(x)=2g(x)g'(x)\ \ \ (*)$$ Thus, if r is a real or complex root of $g(x)$ then either $T'(f(r))=0$ or $f'(r)=0$ .

Since $\deg f(x) < \deg g(x)$ , there exists an either real or complex number $\alpha$ satisfying $g(\alpha) = 0$ and $f(\alpha) \ne 0$ . In other words, from (*) we infer that, there must be one root $r$ of $g(x)$ which is also the root of $T(f'(x)$ .

Also, note that $T'(x)=3x^2-24x+45=0 \Leftrightarrow x \in \{ 3;5\}$ .Therefore, $f(r) \in \{3;5\} \Rightarrow P(3)=0 \vee P(5)=0 \Leftrightarrow 54-a=0 \vee 50-1=0$

which is why $a \in \{50;54\}$ and so the answer is 104.

Note first that the RHS can be rewritten as $\left[(f(x))^3-12(f(x))^2+48f(x)-64\right]-3f(x)-(a-64)$ which can then be simplified to $(f(x)-4)^3-3f(x)-(a-64)$ . Let $h(x)=f(x)-4$ . Then the equation reduces to $(g(x))^2=(h(x))^3-3(h(x)+4)-(a-64)$ $(g(x))^2=(h(x))^3-3h(x)+(a-52).$ There are several things we can notice about the fact that the RHS must be a perfect square polynomial. First, note that $g(x)$ can not be an irreducible poly-polynomial in terms of $h(x)$ (for example, $(h(x))^2+h(x)+1$ ). To prove this, note that the degree of the RHS is $3$ , which is an odd number. If $g(x)$ were irreducible, then $(g(x))^2$ would have an even degree, contradiction. Therefore $(g(x))^2$ must be broken down into linear factors.

Because of this, let $(h(x))^3-3h(x)+(a-52)=(h(x)-a_1)(h(x)-a_2)(h(x)-a_3)$ for some real $a_1,a_2,a_3$ . We must next note that it is impossible for all three of these factors to be perfect square polynomials. To prove this, suppose the opposite, that $a(x)$ and $b(x)$ are both perfect square polynomials that differ by a constant. Then the constant terms in $\sqrt{a(x)}$ and $\sqrt{b(x)}$ would differ. But note that due to the distributive property, the constant terms of each of these polynomials affect other coefficients as well, which is a direct contradiction of the fact that they only differ in their constant term. Therefore only one of the three linear factors can be a perfect square polynomial. It follows that two of the $a_i$ must be equal.

Because of this, rewrite our original equation as $(h(x))^3-3h(x)+(a-52)=(h(x)-a_1)^2(h(x)-a_2).$ It is our job now to determine the possible values for $a_1$ and $a_2$ . Consider this as a polynomial in terms of $h(x)$ , with roots $a_1,a_1,a_2$ . By Vieta's formulas, we must have $a_1+a_1+a_2=0\implies a_2=-2a_1$ . Then we can use Vieta's one more time to get that $a_1a_1+a_1a_2+a_1a_2=-3$ $a_1^2-2a_1^2-2a_1^2=-3$ $-3a_1^2=3$ $a_1=\pm 1$

By substituting to find $a_2$ , we see that the only possible factorizations of $(h(x))^3-3h(x)+(a-52)$ that result in perfect square trinomials are $(h(x)+1)^2(h(x)-2)$ and $(h(x)-1)^2(h(x)+2)$ . We can check to see that suitable $h(x)$ indeed produce valid $g(x)$ . (For example, $h(x)=x^2+2x+3$ satisfies the first case, and $h(x)=x^2+2x-1$ satisfies the second case.) The constant terms in these polynomials are $-2$ and $2$ , respectively. Letting $a-52=-2$ gives $a=54$ , and letting $a-52=2$ gives $a=50$ . Hence the only two possible values of $a$ are $54$ and $50$ , and the sum of these two values is $\boxed{104}$ .

Lemma: For $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \in \mathbb{R}$ such that $\alpha_{1} \neq \alpha_{2} \neq \ldots \neq \alpha_{n}$ and $p(z), q(z)$ any two polynomials over $\mathbb{R}$ : $(p(z))^2=\prod_{i=1}^{n} (q(z)-\alpha_{i}) \Rightarrow p(z), q(z)$ are constant (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 $\alpha_{1} \neq \alpha_{2} \neq \ldots \neq \alpha_{n}$ , the set of roots for $q(z)-\alpha_i,i=1, \ldots,n$ are disjoint from each other. Since the roots of the above equation come in pairs and $\mathbb{R}$ is a 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: $p(z)= \prod_{i=1}^{n} p_i (z)$ and $(p_i (z))^2=q(z)-\alpha_i, i=1,2, \ldots, n$ Notice that for $1 \leq i \neq j \leq n$ , $(p_i (z) +p_j (z))(p_i (z) -p_j (z))=(q(z)-\alpha_i)-(q(z)-\alpha_j)=\alpha_j-\alpha_i \neq 0$ are constants. Thiis implies all $p_i (z)$ are constants. As a result, $p(z)=\prod_{i=1}^{10} p_i (z)$ and $q(z)=(p_i (z))^2+\alpha_i$ are also constants, a contradiction!

Using the lemma to the problem, $(g(x))^2=P(f(x))=(f(x))^3-12(f(x))^2+45f(x)-a$ , for it to have non-constant solutions, a necessary condition is that 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 $P'(c)=3c2-24c+45=3(c-3)(c-5)=0\Rightarrow$ $c=3$ or $c=5$ . $P(3)=0$ gives $a=54$ and $P(5)=0$ gives $a=50$ . Therefore, the answer is $50+54=104$ .

Claim: For $a_{1}, a_{2}, \ldots, a_{n} \in \mathbb{R}$ such that $a_{1} \neq a_{2} \neq \ldots \neq a_{n}$ and $p(z), q(z)$ any two polynomials over $\mathbb{R}$ : $(p(z))^2=\prod_{i=1}^{n} (q(z)-a_{i}) \Rightarrow p(z), q(z)$ are constant Proof: Assume the contrary, suppose there are two non-constant polynomials $p(z),q(z)$ such that $(p(z))^2=\prod_{i=1}^{n} (q(z)-a_{i})$ . Since $a_{1} \neq a_{2} \neq \ldots \neq a_{n}$ , the set of roots for $q(z)-a_i,i=1, \ldots,n$ are disjoint from each other. Since the roots of the above equation come in pairs and $\mathbb{R}$ is a Unique Factorization Domain, the roots of each $q(z)-a_i$ also come in pairs. This means we can factor $p(z)$ into n pieces such that: $p(z)= \prod_{i=1}^{n} p_i (z)$ and $(p_i (z))^2=q(z)-a_i, i=1,2, \ldots, n$ Notice that for $1 \leq i \neq j \leq n$ , $(p_i (z) +p_j (z))(p_i (z) -p_j (z))=(q(z)-a_i)-(q(z)-a_j)=a_j-a_i \neq 0$ are constants. Thiis implies all $p_i (z)$ are constants. As a result, $p(z)=\prod_{i=1}^{10} p_i (z)$ and $q(z)=(p_i (z))^2+a_i$ are also constants, a contradiction!

Applying the claim to the original problem, $(g(x))^2=P(f(x))=(f(x))^3-12(f(x))^2+45f(x)-a$ , for it to have non-constant solutions, a necessary condition is that 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 $P'(c)=3c2-24c+45=3(c-3)(c-5)=0\Rightarrow$ $c=3$ or $c=5$ . $P(3)=0$ gives $a=54$ and $P(5)=0$ gives $a=50$ . Hence, the answer is $50+54=104$ .

Let $P(x)=x^3-12x^2+45x$ , we have $P’(x) = 0$ when $x = 3$ or $x=5$ . Note that $P(3)=54$ and $P(5)=50$ . 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 for $a<50, a>54$ , 3 distinct real roots for $50<a<54$ , and 2 real roots (with 1 repeated) for $a=50, 54$ . Assume that $a \not =50, 54$ . We will show that 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 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 $h(x) \in \mathbb{R}[x]$ . 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.

We can check directly with $a=50, a = 54$ . Therefore, the sum we want to find is $50+54=104$ .