Tri-ratio

Let... $f(x)=\dfrac{x^b+x-1}{x^a+x^2-1}$

We can rewrite this in the form... $x^b+x-1=\left(x^a+x^2-1\right)\cdot f(x)$

So we have... $f(x)=\sum_{i=0}^{b-a} p_{b-a-i} x^{b-a-i} ~~~~~~~~~~ ~ [p_{b-a}=1]$

Now keep noting... :p

$\left(x^a+x^2-1\right)\mid \left(x^b+x-1\right)$
$\because ~~~ x^b+x-1=x^{b-a}\left(x^a+x^2-1\right)+\left(1-x\right)\left(x^{b-a+1}+x^{b-a}-1\right)$

$\therefore ~~~ \left(x^a+x^2-1\right)\mid \left(x^{b-a+1}+x^{b-a}-1\right)$

$\therefore ~~~ \deg{\left(x^a+x^2-1\right)}\le \deg{\left(x^{b-a+1}+x^{b-a}-1\right)}\Longrightarrow b \geq 2a-1$

$\because ~~~ \left(x^a+x^2-1\right) \mid \left(\left(x^{b-a+1}+x^{b-a}-1\right)-x^{b-2a+1}\left(x^a+x^2-1\right)\right)$

$\therefore ~~~ \left(x^a+x^2-1\right) \mid \left(x^{b-a}-x^{b-2a+3}+x^{b-2a+1}-1\right):=h(x)$

$\therefore ~~~ h(x)=k\left(x^a+x^2-1\right)$

Note that, $x^a+x^2-1$ is equal to $1$ at $x=1$ and $-1$ at $x=0$ ... Hence, it must have a root in the interval $(0,1)$ (if this is the exclusive interval notation, I ain't sure :p)...

• Case 1 : $b=2a-1$ ... So we have $h(x)=x^{a-1}-x^2$ ... For $a=3$ , $h(x)=0$ and suffices the conditions, and also yields $b=5$ ... For $a>3$ , $h(x)=x^2\left(x^{a-3}-1\right)$ ... But this has no root in the interval $(0,1)$ , hence, $\left(x^a+x^2-1\right)$ can't be a factor of $h(x)$ ... So we're left with exactly $1$ solution from this case: $(a,b)=(3,5)$ ...

$~$

• Case 2 : $b>2a-1$ ... But then $h(x)$ factorizes as the following, and so it has no roots in the interval $(0,1)$ implying that $\left(x^a+x^2-1\right)$ can't be a factor of $h(x)$ ...

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

$=\left(x^{b-a}-1\right)-\left(x^{b-2a+3}-x^{b-2a+1}\right)$

$=\left(x-1\right) \left(\left(\sum_{i=1}^{b-a}x^{m-n-i}\right) -x^{b-2a+2}-x^{b-2a+1}\right)$

It's trivial to check that $(a,b)=(1,2)$ is another solution, observing the expression...

In total, we have the sum of the values of $a$ : $3+1=\fbox{4}$

Note : This is a very slightly modified version of Problem 3 from IMO 2002 ... I once solved it in this elementary method...

*Note: * This problem has been adapted from IMO 2002 . I took some of the ideas from there to complete my solution.

The case $a=1$ will be dealt with separately. For the time being, we let $a \geq 2$ .

Let $\dfrac{x^b+x-1}{x^a+x^2-1}= P(x)$ . We can rewrite this as $x^b+x-1= P(x)(x^a+x^2-1)$ It then immediately follows that if $C$ is a root of $x^a+x^2-1$ , it must also be a root of $x^b+x-1$ . Let $x^a+x^2-1= f_a(x)$ . Note that $f_a(0)= -1<0$ and $f_a(1)= 1>1$ . From IVT, we conclude $f_a(x)$ must have a root $C$ in the range $(0, 1)$ . This must also be a root of $x^b+x-1$ , so $C^b+C-1=0 \implies C^b+C= 1$ Again, $C$ is a root of $x^a+x^2-1$ , so $C^a+C^2-1= 0 \implies C^a+C^2= 1$ Combining them, we find out $C^a+C^2= C^b+C$ $\implies C(C-1)= C^a \left ( 1 - C^{b-a} \right )$ We know that $C^a + C^2 - 1= 0$ So $C^a= 1-C^2$ $\implies C(C-1)= (1-C^2) \left ( 1 - C^{b-a} \right )$ Since $C \neq 1$ , we can divide both sides by $C-1$ to obtain: $C= (C+ 1) \left ( 1 - C^{b-a} \right )$ $\implies C= C - C^{b-a+1} + 1 - C^{b-a}$ $\implies C^{b-a+1} + C^{b-a} = 1$ Since $0 < C < 1$ , $C^m > C^n$ for all $n>m>0$ . It then follows that if $b-a+1 \geq a$ , $C^{b-a+1} + C^{b-a} < C^{a} + C^{a-1} \leq C^a + C =1$ Where in the last step we used the assumption $a\geq 2 \implies a-1 \geq 1$ . This is a contradiction, so $a \leq b-a+1$ .

Now, note that $x^b+x-1= P(x)(x^a+x^2-1)$ $\implies x^{b-a} \left ( x^a + x^2 -1 \right ) - \left ( x^b + x - 1 \right ) = \left ( x^{b-a} - P(x) \right ) (x^a + x^2 - 1)$ $\implies x^{b-a+2} - x^{b-a} -x + 1 = \left ( x^{b-a} - P(x) \right ) (x^a + x^2 - 1)$ We then conclude $x^a+x^2-1 \mid x^{b-a+2} - x^{b-a} -x + 1$

The degree of $x^{b-a+2} - x^{b-a} -x + 1$ is $b-a+2$ , which has to be $\geq a$ . But since $b-a+1 \leq a$ , we have two possibilities:

• $b-a+2= b \implies a=2$

• $b-a+2 = b+1 \implies a= 3$

*Case $1$ : * $a=2$
In this case, $f_a(x)= 2x^2-1$ , which has roots $\pm \dfrac{1}{\sqrt{2}}$ . These must also be the roots of $x^b+x-1$ , so $\left ( \frac{1}{\sqrt{2}} \right ) ^b = 1 - \frac{1}{\sqrt{2}}$ $\left ( -\frac{1}{\sqrt{2}} \right ) ^b = 1 + \frac{1}{\sqrt{2}}$ It can be verified that none of these produce integer solutions for $b$ .

*Case $2$ : * $a=3$
We can verify $b=5$ satisfies the conditions, so $a=3$ is a solution.

We also have the trivial solution $(a,b)= (1,1)$ . The sum of all possible values of $a$ , thus, is $3+1= \boxed{4}$ .

I simply assumed that for $a=1$ there will be at least one polynomial in the form given divisible by that expression. Then, I found $b=5$ , $a=3$ is another valid solution. Thus, $3+1=\boxed{4}$ . This is NOT a rigorous solution, I was just playing around.