Fancy Functions

Represent f(x) as f, and represent g(x) as g.

Obviously, if two polynomials are equal for infinite values, the polynomials must be equal (following from remainder and factor theorem), thus we have:

f^3 + af + b = g^3 + cg^2

It is clear f and g must have the same degree. We now aim to show that f-g is a constant.

(f-g)(f^2 + fg + g^2) = cg^2 - af + b Let the coefficient of the largest power of x in f and g be M and N respectively (M^2 + MN + N^2) = (1/2)((M+N)^2 + M^2 + N^2) > 0 since M and N are non zero.
Thus (f-g) must have degree 0 since the degree of f^2 + fg + g^2 is at least equal to the degree of cg^2, thus f - g is a constant.

Thus let f = g + k for some constant k

(f-g)(f^2 + fg + g^2) = cg^2 + af + b

(k)(3g^2+3gk+k^2) = cg^2 - ag - ak + b

Which becomes a polynomial in g.

Comparing coefficients of g^2, g^1 and g^0, we have:

3k = c

3k^2 = -a

k^3 = -ak + b

Simple manipulation of the second and third equations yields:

b = -2k^3, c = 3k

Since b = -2k^3 and c = 3k must both be integer, this implies that k can only take the values of (-1,-2,-3...,-7) s.t. b stays with the bound of (1,1000)

(to see this, let k = -i/3 for some integer i. -2k^3 = (2i^3)/27 is integer implying i is a multiple of 3, implying k itself is a multiple of 3)

Application of Remainder and factor theorem: If two polynomials P and Q are equal for infinite values, the polynomial P-Q is zero for infinite values - this means that these infinite values are all roots, meaning that P-Q must be the zero polynomial, thus P = Q.

Let $m= \deg (f(x))$ and $n= \deg (g(x))$ . Note that $\deg ((f(x))^3 + af(x) + b)= 3m$ $\deg ((g(x))^3 + c(g(x))^2)= 3n$ Equating them, $3m= 3n \implies m=n$ We now rewrite the equation: $(f(x))^3 - (g(x))^3= c(g(x))^2 - af(x) - b$ This factorizes: $(f(x)-g(x))(f(x)^2 + f(x)g(x) + g(x)^2)= c(g(x))^2 - af(x) + b$ Let $p$ and $q$ denote the leading coefficients of $f(x)$ and $g(x)$ respectively. Note that $p^2 + pq + q^2 = \frac{1}{2} ((p+q)^2+p^2+q^2) > 0$ The degree of $f(x)^2 + f(x)g(x) + g(x)^2$ , thus, is $\max (\deg (f(x)^2, \deg (f(x)g(x), \deg (g(x)^2)$ , i.e. $\max (2m, m+n, 2n)$ . This is definitely greater than or equal to the degree of $c(g(x))^2 - af(x) + b$ , which is $\geq \max (2n, m)$ . We want equality to occur. Also note that if $\deg (f(x)-g(x)) > 0$ , the degree of the LHS is clearly greater than the degree of the RHS. So, $f(x)-g(x)$ is a constant. Let $f(x)-g(x)=k$ . Substituting $f(x)=g(x)+k$ , we find out $k(3g(x)^2 + 3kg(x) + k^2)= cg(x)^2 - ag(x)+-ak +b$ We can consider this as a polynomial equation in $g(x)$ . Let's see the coefficients of $g(x)^0, g(x), g(x)^2$ in the LHS and RHS. $\begin{array}{c|c|c} \text{Term} & \text{Coefficient in LHS} & \text{coefficient in RHS} \\ \hline g(x)^0 & k^3 & b-ak \\ \hline g(x) & 3k^2 & -a \\ \hline g(x)^2 & 3k & c \\ \hline \end{array}$ Equating them, we find out $c= 3k$ and $b= -3k^3$ . The first equation means $k= \dfrac{c}{3}$ , and the second equation means $2k^3= -\dfrac{c^3}{27}$ . It can be checked that the set of values of $k$ which keeps $b$ in the range $[1, 1000]$ is $\{-1, -2, \cdots , -7 \}$ , which shows that there are $\boxed{7}$ values of $b$ in all.

It's clear to show that $f(x)$ has the same the leading coefficient as $g(x)$ has.

Suppose that: $deg f=deg g=n,n\ge 1$ . Let: $h(x)=g(x)-f(x)$ , we get $deg h=r<n$ .

From the hypothesis, we have:

$f^3(x)+af(x)+b=(f(x)+h(x))^3+c(f(x)+h(x))^2$

or $3f^2(x)h(x)+3f(x)h^2(x)+h^3(x)+c(f(x)+h(x))^2-af(x)-b=0$ .

If $r\geq 1$ , degree of the polynomial in the left side is $2n+r$ , , a contradiction.

If $r=0$ , assume that: $h(x)\equiv k$ .

Hence, $f^3(x)+af(x)+b=(f(x)+k)^3+c(f(x)+k)^2$

or $(c+3k)f^2(x)+(2ck+3k^2-a)f(x)+ck^2+k^3-b=0$ , for all $x$ .

This implies that:

$\begin{cases}c+3k=0\\2ck+3k^2-a=0\\ck^2+k^3-b=0\end{cases} \Leftrightarrow \begin{cases}c=-3k\\a=-3k^2\\b=-2k^3\end{cases}$

Because $a,b,c$ are integer numbers and $0<b<1000$ , $b\in\{2;16;54;128;250;432;686\}$ . There are 7 values of $b$ satisfy.

Since $f(x)^3 + af(x) + b \,=\, g(x)^3 + cg(x)^2$ , we deduce that $f(x)$ and $g(x)$ must have the same degree $n \in \mathbb{N}$ .

Since $af(x) + b$ has degree $n$ and $cg(x)^2$ has degree $2n$ , the coefficients of $x^r$ in $f(x)^3$ and $g(x)^3$ must be the same for $2n+1 \le r \le 3n$ . If $f(x) \; = \; \sum_{j=0}^n a_j x^j \qquad g(x) \; = \; \sum_{j=0}^n b_j x^j$ we deduce that $\sum_{u+v+w=2n+r}a_ua_va_w \; = \; \sum_{u+v+w=2n+r}b_ub_vb_w \quad,\quad 1 \le r \le n$ When $r=n$ we deduce that $a_n^3 = b_n^3$ , and hence $a_n = b_n \neq 0$ . When $r=n-1$ we deduce that $3a_n^2a_{n-1} = 3b_n^2b_{n-1}$ , and hence $a_{n-1} = b_{n-1}$ . Suppose that $2 \le k \le n-1$ and that $a_{n-j} = b_{n-j}$ for all $0 \le j \le k-1$ . For every $u,v,w$ such that $u+v+w = 3n-k$ , then either $u,v,w \ge n+1-k$ , and hence $a_ua_va_w = b_ub_vb_w$ , or else one of the coefficients $u,v,w$ is equal to $n-k$ , with the other two equal to $n$ . This implies that $\sum_{u+v+w=2n+r}a_ua_va_w \; = \; 3a_n^2a_{n-k} + F \qquad \sum_{u+v+w=2n+r}b_ub_vb_w \; = \; 3b_n^2b_{n-k} + F$ for some constant $F$ . Thus $3a_n^2a_{n-k} = 3b_n^2b_{n-k}$ , and hence $a_{n-k} = b_{n-k}$ . Thus we deduce that $a_j = b_j$ for all $1 \le j \le n$ , so that $f(x) = g(x) + \alpha$ for some constant $\alpha$ . But then $\begin{array}{rcl} g(x)^3 + cg(x)^2 & =& f(x)^3 + af(x) + b \\ & = & g(x)^3 + 3\alpha g(x)^2 + (3\alpha^2 + a)g(x) + (\alpha^3 +a\alpha + b) \end{array}$ which implies that $c = 3\alpha$ , $3\alpha^2+a = 0$ and $\alpha^3 + a\alpha + b = 0$ . These equations imply that $a=-3\alpha^2\,,\,b=2\alpha^3\,,\,c=3\alpha$ . Since $a,b,c$ are to be integers, with $b$ positive, we deduce that $\alpha$ must be a positive integer; since $b < 1000$ we deduce that $1 \le \alpha \le 7$ .

Thus there are $7$ possible values of $b$ . This problem is basically one of carefully matching the coefficients in the identity, and making as much use as possible of the $n$ equations we get which do not involve the complicating integers $a,b,c$ .

Given equation is $f^3(x)+af(x)+b=g^2(x)(g(x)+c)$ . Let the equation $f^3(x)+af(x)+b$ has two equal roots (let $\alpha,\alpha$ ) and the other root is $\beta$ .

From the equation we get, $2\alpha+\beta=0 or, 2\alpha=-\beta, \alpha^2+\beta=-b$ and $\alpha^2+2\alpha \beta =a$ .

As $b$ is positive, $\beta<0$ and $\alpha>0$ .

Now, by some simple calculation we get, $b=2\alpha^3$ and $a=-3\alpha^2$ .

As $b<1000$ , the possible values of $\alpha$ are $1,2,3,4,5,6,7$ . So, the answer is $\boxed{7}$