Polynomials and squares

Since $f(x)$ is monic, we can write it as $x^2+bx+c$

Thus we can write $g(x)$ as

$b^2x^2+2bcx+2bx^3+c^2+2cx^2+x^4-(x^4+bx^2+c)$

$=b^2x^2+2bcx+2bx^3+c^2+2cx^2-bx^4-c$

$=2bx^3+(b^2-b+2c)x^2+2bcx+c^2-c$

Since $g(x)$ is monic, we know that $2b=1$ , therefore $b=\frac{1}{2}$

Now $g(x)$ can be further simplified as $x^3+(2c-\frac{1}{4})x^2+cx+c^2-c$

It is now easy to see that for $g(x)$ to have three non-zero coefficients, either $2c-\frac{1}{4}=0$

or $c^2-c=0$

We find values of $\frac{1}{8}$ and $1$ for $c$

So $f(x)=x^2+\frac{1}{2}x+\frac{1}{8}$ and $x^2+\frac{1}{2}x+1$ are the satisfying quadratics

Which yield a $f(0)$ value of $\frac{1}{8}$ and $1$ respectively

Thus the sum of the $f(0)$ values is $\frac{9}{8}$ .

So we get $m+n=\boxed {17}$

First off, let us define the polynomials.

$f(x)=x^{2}+ax+b$

$f(x^{2})=x^{4}+ax^{2}+b$

$(f(x))^{2}=x^{4}+2ax^{3}+(a^{2}+2b)x^{2}+2abx+b^{2}$

Now this must mean that

$g(x)=x^{4}+2ax^{3}+(a^{2}+2b)x^{2}+2abx+b^{2}-x^{4}+ax^{2}+b=$

$=2ax^{3}+(a^{2}-a+2b)x^{2}+2abx+b(b-1)$

Now, considering $g(x)$ is also monic, we can conclude that its leading coefficient $2a$ equals $1$ , so we can easily find $a$

$2a=1 \Rightarrow a=\frac{1}{2}$

Thus, $g(x)$ takes a different form

$g(x)=x^{3}+(2b-\frac{1}{4})x^{2}+bx+b(b-1)$

Also, since $g(x)$ has exactly $3$ non-zero coefficients and one of them is already $1$ , then that leaves us with $2$ more. Since there are $3$ coefficients left in the polynomial, this must mean that one of them equals $0$ . With that said, we may observe $3$ different cases.

Case 1:

$b(b-1)=0$

From this case, we may derive $2$ more sub-cases:

• $b=0$ , for which case $f(x)=x^{2}+\frac{1}{2}x+0$
• $b-1=0 \Rightarrow b=1$ , for which case $f(x)=x^{2}+\frac{1}{2}x+1$

Case 2:

$b=0$

We have already discussed this instance, so again, $f(x)=x^{2}+\frac{1}{2}x+0$ for this one.

Case 3:

$2b-\frac{1}{4}=0 \Rightarrow 2b=\frac{1}{4} \Rightarrow b=\frac{1}{8}$

For this case, $f(x)=x^{2}+\frac{1}{2}x+\frac{1}{8}$ .

Now that we have found all possible $f(x)$ polynomials with these properties, we may find the sum $S$ of all such $f(0)$ polynomials, giving us

$S=0+0+0+0+0+1+0+0+0+0+0+\frac{1}{8}=\frac{9}{8}$

Since we know that $\frac{m}{n}=\frac{9}{8}$ , we can see that $m+n=17$ .

Therefore, the answer is $17$ .

$f(x) = x^2 + ax + b \\ \Downarrow \\ g(x) = (2a)x^3 + \left( a(a-1)+2b \right) x^2 + (2ab)x + b(b-1)$

We know that $g(x)$ is a monic polynomial, so $2a=1$ , implying

$g(x) = x^3 + \left( 2b - \frac{1}{4} \right) x^2 + bx + b(b-1).$

Exactly one of these coefficients equals $0$ . However,

$b = 0 \implies b(b-1) = 0,$

so either

$\left( 2b - \frac{1}{4} \right) = 0 \implies f(0) = b = \frac{1}{8}$

or

$b(b-1) = 0 \implies f(0) = b = 1.$

Hence the sum of $f(0)$ for all such quadratic polynomials can be written as

$\frac{m}{n} = \frac{9}{8},$ so $m + n = 9 + 8 = \boxed{17}.$

Since $f(x)$ is monic and quadratic , we can generalize that $f(x) = x^{2} + ax + b$

$g(x) = (f(x))^{2} - f(x^{2})$

$g(x) = (x^{2}+ax+b)^{2} - ((x^{2})^{2}+a(x^{2}) +b)$

$g(x) = ((x^{2}+ax)+b)^{2} - (x^{4}+ax^{2} +b)$

$g(x) = ((x^{2}+ax)^{2}+2b(x^{2}+ax)+b^{2})-(x^{4}+ax^{2}+b)$

$g(x) = x^{4} + 2ax^{3} + a^{2}x^{2}+2bx^{2}+2abx+c^{2}-x^{4}-ax^{2}-c$

Simplifying we get

$g(x) = 2ax^{3}+(a^{2}+2b-a)x^{2}+2abx+(b^{2}-b)$

Since $g(x)$ is monic , it means the coefficient of $x^{3}$ is 1. hence $2a=1$ or $a=\frac{1}{2}$

$g(x) = x^{3}+(2b-\frac{1}{4})x^{2}+bx+(b^{2}-b)$

now $g(x)$ only has three non-zero coefficients, which means that either $2b-\frac{1}{4}=0$ or $b=0$ or $b^{2}-b=0$

if $b=0$

$g(x) = x^{3}-\frac{1}{4}x^{2}$

this does not satisfy as there are only 2 non-zero coefficients

if $b^{2}-b=0$ or $b=1$ then

$g(x) = x^{3}+\frac{7}{4}x^{2}+x$

this satisfies the condition of 3 non-zero coefficients. lets try the final one $2b-\frac{1}{4}=0$ or $b=\frac{1}{8}$

$g(x) = x^{3}+\frac{1}{8}x-\frac{7}{64}$

this also satisfies the condition of three non-zero terms. hence the two values of $b$ are $\frac{1}{8}$ and $1$ .

lets sub this into our original $f(x)$

$f(x)=x^{3}+ax+1$ or $f(x)=x^{3}+ax+\frac{1}{8}$

$f(0) = 1$ or $f(0) = \frac{1}{8}$

hence the sum of $f(0)$ for all the possible quadratic polynomials is $1+\frac{1}{8} = \frac{9}{8}\Rightarrow 9+8=17$

Let,

\begin{equation} f(x) = x^{2} + c 1x + c 2 \end{equation}

where $c_1$ and $c_2$ are some real constants. The $x^2$ terms cannot have a constant as it must be a monic quadratic. Using this definition of $f(x)$ we can determine a function for $g(x)$ ,

\begin{equation} g(x) = (f(x))^{2}-f(x^{2}) \end{equation}

\begin{equation} g(x) = (x^{2} + c 1x + c 2)^{2}-((x^{2})^{2} + c 1x^{2} + c 2) \end{equation}

Expanding and simplifying this expression results in,

\begin{equation} g(x) = 2c 1x^{3} + (2c^{2}+c 1(c 1-1))x^{2}+2c 1c 2x+c 2(c_2-1) \end{equation}

We know from the question that $g(x)$ must also be a monic polynomial and therefore the leading coefficient must be equal to one. It also must have three non-zero coefficients. For there to be three non-zero coefficients, $g(x)$ must be either a monic cubic polynomial with one coefficient being a zero or a quadratic polynomial with no coefficients being zero. Assuming $g(x)$ is a monic cubic polynomial,

\begin{equation} 2c_1 = 1 \end{equation}

\begin{equation} c_1 = \frac{1}{2} \end{equation}

Simplifying $g(x)$ ,

\begin{equation} g(x) = x^{3}+(2c 2-\frac{1}{4})x^{2}+c 2x+c 2(c 2-1) \end{equation}

For there to be exactly three non-zero coefficients either the $x^{2}$ term, $x$ term or the constant term has to be equal to zero. Setting each coefficient equal to zero results in,

$c_2 = \frac{1}{8} \quad$ or $\quad c_2 = 0 \quad$ or $\quad c_2 = 1$

When substituting these values of $c_2$ back into $g(x)$ we realize that when if $c_2$ is zero that two coefficients become zero, making the total number of non-zero coefficients only two. This does not satisfy the terms of the question and therefore,

$c_2 = \frac{1}{8} \quad$ or $\quad c_2 = 1$

Now we must consider if $g(x)$ is a monic quadratic polynomial. For $g(x)$ to be a quadratic,

\begin{equation} 2c_1 = 0 \end{equation}

\begin{equation} c_1 = 0 \end{equation}

But if $c_1$ is equal to 0, the $x$ term of $g(x)$ also become zero making it only a two term polynomial making it impossible to have three non-zero coefficients. Therefore it is impossible for $g(x)$ to be a monic quadratic polynomial. Now we know that,

\begin{equation} f(x) = x^{2}+\frac{1}{2}x+\frac{1}{8} \end{equation}

or

\begin{equation} f(x) = x^{2}+\frac{1}{2}x+1 \end{equation}

By substituting zero into both of these expressions we determine that,

\begin{equation} f(0) = \frac{1}{8} \end{equation}

or

\begin{equation} f(0) = 1 \end{equation}

Therefore the sum of the possible values of $f(0)$ is,

\begin{equation} \frac{1}{8} + 1 = \frac{m}{n} = \frac{9}{8} \end{equation}

Therefore $m + n$ is,

\begin{equation} m + n = 9 + 8 = \fbox{17} \end{equation}

$f(x)=x^2+bx+c$

$g(x)=(f(x))^2-f(x^2)=2bx^3+(2c+b^2-b)x^2+2bcx+(c^2-c)$

$2b$ must either be $1$ or $0$ for $g(x)$ to be monic. However, if it is $0$ then $2bc$ is also $0$ which means it does not have three non-zero coefficients, therefore the only solution is $2b=1$ and $b=1/2$

Because the first term is $1$ , one of the other three must be $0$ . However, if $2bc$ is $0$ then $c^2-c$ is also $0$ . Therefore $c \not= 0$

$2c+b^2-b = 0$ or $c^2-c = 0$

$c = \frac{1}{8}$ or $c = 0, 1$ but we already know $c \not= 0$

$\frac{1}{8} + 1 = \frac{9}{8}$ so the answer is 17

f(x) is monic so we can take f(x)= x^2 +bx + c and g(x) = {f(x)}^2 - f(x^2) = 2bx^3 + (b^2-b+2c)x^2 +2bcx +c^2-c * *g(x) is monic so either 2b= 0 and (b^2-b+2c) = 1 Or 2b= 1 . b cannot be 0 since that will reduce g(x) to two terms . Hence 2b= 1 or b = \frac{1}{2} Again g(x) has exactly three non-zero coefficients . Therefore one of (b^2-b+2c) , c^2-c or 2bc must be zero clearly c also cannot be 0 hence c= 1 or (b^2-b+2c) =0 i.e c=1 or c = \frac{1}{8} . Now the sum of f(0) is the sum of possible values of c 1+\frac{1}{8} = \frac{9}{8} 9+8=17

Let $f(x) = x^2+ax+b$ . Then $f(0)=b$ . After substitution, we can see that

$g(x) = 2ax^3 + (a^2-a+2b)x^2 + 2abx + (b^2 - b)$ .

Since $g(x)$ is monic, $a=\frac{1}{2}$ . Plugging into $g(x)$ to obtain

$g(x) = x^3 + (2b-\frac{1}{4})x^2 + bx + (b^2 - b)$ .

We also know that $g(x)$ has three nonzero coefficients. Suppose that the coefficient of $x$ is zero, then $b=0$ and thus the constant term will also be zero, a contradition. Thus the zero coefficient must be either the constant term or the coefficient of $x^2$ . This yield only the value of $b=1,\frac{1}{8}$ . Thus the sum of all possible $f(0)$ is $1+\frac{1}{8}=\frac{9}{8}$ and the answer is $9+8=17$

Given that $f(x)$ is a monic polynomial, lets assume that $f(x)=x^2+bx+c$ .

As per the question g(x)=(f(x))^2-f(x^2)=2bx^3+(b^2+2c-b)x^2+2bcx+c^2-c \.

Since $g(x)$ is also monic, we have $2b=1 \Rightarrow b=1/2$ . Hence $g(x)=x^3+\left(2c-\frac{1}{4}\right)x^2+cx+c^2-c$

For $g(x)$ to have exactly three non-zero coefficients, either the coefficient of $x^2$ or coefficient of $x$ or the constant term must be zero. From this, we get three values of $c$ i.e $\displaystyle c=0,1,\frac{1}{8}$ .

As $f(0)=c$ , the sum of different values of $f(0)$ is $\displaystyle 0+1+\frac{1}{8}=\frac{9}{8}$ .

Therefore, the answer is $9+8=\fbox{17}$ .