Shift by one

We say that a polynomial is nice if it can be written as $g(h(x))$ , where $g(x)$ and $h(x)$ are non-linear polynomials.

Step 1 : If $f(x)$ is nice, then so is $f(x-c)$ for any constant $c$ .

Proof : If $f(x) = g(h(x))$ , we have $f(x-c) = g(h(x-c))$ so it suffices to replace $g(x), h(x)$ with $g(x), h(x-c)$ respectively. (QED)

Hence, we may get rid of the $x^5$ term in $f_A(x)$ by substituting $x$ with $x-1$ to obtain:

$\begin{aligned} f_A(x-1) &= (x^6 - 10x^4 - 10x^3 + x^2 - 10x + 28)\\ &+ A(x^4 + x^3 + x - 3) \end{aligned}$

Step 2 : If $f(x)$ is nice and $f(x) = g(h(x))$ for non-linear $g(x), h(x)$ , we may assume with no loss of generality that $h(0) = 0$ and $h(x)$ is monic.

Proof : For any constants $\alpha, \beta (\alpha \ne 0)$ , replace $g(x), h(x)$ by $g((x-\beta)/\alpha), \alpha h(x)+\beta$ respectively. We can choose $\alpha, \beta$ such that $h(x)$ is monic and has no constant term. (QED)

Step 3 : For $f_A(x-1)$ above, if $\deg(g) = 3$ and $\deg(h) = 2$ , then $A = 10$ is the only solution.

Proof : By step 2, write $h(x) = x^2 + bx$ . Since the coefficient of $x^5$ is zero, we have $b=0$ as well. Hence, $h(x) = x^2$ and all the odd terms of $f_A(x-1)$ vanish, i.e. $A=10$ . Conversely, for $A=10$ , all odd terms vanish so $f_A(x-1)$ is a polynomial in $x^2$ . (QED)

Step 4 : For $f_A(x-1)$ above, if $\deg(g) = 2$ and $\deg(h) = 3$ , then $A = 12$ is the only solution.

Proof : Write $h(x) = x^3 + bx^2 + cx$ by step 2. Once again, by step 1, $b=0$ . So we have:

$f_A(x-1) = p(x^3 + cx)^2 + q(x^3 + cx) + r$

for some quadratic $g(x) = px^2 + qx + r$ . Comparing the coefficients, we get:

$p=1,\ 2pc = A-10,\ pc^2 = 1,\ q=A-10,\ qc = A-10.$

The last two equations give us $q=qc$ . If $q=0$ , then $A=10$ as in step 3. Otherwise, $c=1$ and we have $A=12$ , in which case we can write $f_A(x-1)$ as:

$\begin{aligned} & x^6 + 2x^4 + 2x^3 + x^2 + 2x + 8 \\ = & (x^3 + x)^2 + 2(x^3 + x) + 8. \end{aligned}$

Since $g(h(0))=0$ , we note that, replacing $g(x)$ and $h(x)$ by $\hat{g}(x) = g(x+h(0))$ and $\hat{h}(x) = h(x) - h(0)$ , we may assume without loss of generality that $g(0) = h(0) = 0$ . Since $g(h(x))$ is sextic and neither $g(x)$ nor $h(x)$ is linear, we deduce that one of $g(x)$ and $h(x)$ is quadratic, and the other is cubic. We may also assume that $g(x)$ is monic, since any leading coefficient for $g(x)$ can be incorporated into $h(x)$ .

1. If $g(x)$ is quadratic, write $g(x) = x^2 + ax$ , $h(x) = \alpha x^3 + \beta x^2 + \gamma x$ . Multiplying out $g(h(x))$ and matching coefficients, we deduce that $\begin{array}{rclrcl} \alpha^2 & = & 1 & 2\alpha\beta &=& 6 \\ \beta^2 + 2\alpha\gamma &=& A+5 & 2\beta\gamma + a\alpha &=& 5A-30 \\ \gamma^2 + a\beta &=& 9A - 74 & a\gamma & = & 8A-72 \end{array}$ These equations can be solved to obtain $A=12$ and $\alpha^2 = 1 \qquad \beta =3\alpha \qquad \gamma = 4\alpha \qquad a = 6\alpha$

2. If $g(x)$ is cubic, write $g(x) = x^3 + ax^2 + bx$ , $h(x) = \alpha x^2 + \beta x$ . Multiplying out $g(h(x))$ and matching coefficients, we deduce that $\begin{array}{rclrcl} \alpha^3 & = & 1 & 3\alpha^2\beta & = & 6 \\ 3\alpha\beta^2 + a\alpha^2 & = & A+5 & \beta^3 + 2a\alpha\beta & = & 5A-30 \\ a\beta^2 + b\alpha & = & 9A-74 & b\beta & = & 8A-72 \end{array}$ These equations can be solved to obtain $A=10$ and $\alpha^3 = 1 \qquad \beta = 2\alpha \qquad a = 3\alpha \qquad b = 4\alpha^2$

Thus the only possible values of $A$ are $12$ and $10$ , and $12+10 = 22$ .