Find the minimal $a+b$.

Question

Let $x$ be real number such that $\frac{(1+x)^3}{1+x^3} = \frac{9}{13}$. If $\frac{(1+x)^5}{1+x^5} = \frac{a}{b}$ where $a, b$ are positive integers. What is the minimal value of $a+b$.

My attempted work :

$\frac{(1+x^3)(1+x)^2}{(1+x)^5} = \frac{13}{9}$

$\frac{x^5+2x^4+x^3+x^2+2x+1}{(1+x)^5} = \frac{13}{9}$ ---[1]

Since $13(1+x)^3 = 9(1+x^3)$

$13x^3+39x^2+39x+13 = 9+9x^3$

$4x^3+39x^2+39x+4 = 0$ ---[2]

$(x+1)(4x^2+35x+4) = 0$

$(4x^2+35x+4) = 0$

$(x+1)^2 = -\frac{27}{4}x$

so $x = -\frac{4}{27}(x+1)^2$

[2] : multiplied through by $x$, we have

$4x^4+2x^3+2x^2+4x = -37x^3 -37x^2$ ---[3]

[2] : $4x^3+4 = -39x^2-39x$

multiplied through by $\frac{37}{39}x$, we have

$\frac{148}{39}x^4 + \frac{148}{39}x = -37x^3 -37x^2$ ---[4]

[3]=[4] : $4x^4+2x^3+2x^2+4x = \frac{148}{39}x^4 + \frac{148}{39}x = \frac{148}{39}x(x^3+1) = \frac{148}{39}x\left(\frac{13}{9}(1+x)^3\right) = \frac{148}{39}\left(-\frac{4}{27}(x+1)^2\right)\left(\frac{13}{9}(1+x)^3\right) = -\frac{148\cdot4\cdot13}{39\cdot27\cdot9}(x+1)^5 $

so $\frac{2x^4+x^3+x^2+2x}{(x+1)^5} = -\frac{148\cdot2\cdot13}{39\cdot27\cdot9}$ ---[5]

[1]-[5] : we have $\frac{1+x^5}{(1+x)^5} = \frac{13}{9} + \frac{148\cdot2\cdot13}{39\cdot27\cdot9}$

$\frac{(1+x)^5}{1+x^5} = \frac{729}{1349} = \frac{a}{b}$

$a+b = 729+1349= 2078$

Does minimal value in this problem mean gcd of numerator and denominator is equal to 1 ?

Is my answer correct ?

Is there an easier way to find the answer, please suggest ?

Thank you.

I understand everything except the last line. Can you explain how you got there? — uniquesolution, Mar 09 '17 at 15:07
I think a further argument is required to prove that $2078$ is actually the minimal value. — uniquesolution, Mar 09 '17 at 15:25

mathlove · Accepted Answer · 2017-03-10T04:22:56.410

Your answer is correct.

Is there an easier way to find the answer, please suggest ?

How about the following way?

You already have $$4x^2+35x+4=0$$ from which we get $$x+\frac 1x=\frac{-35}{4}\tag1$$

Now dividing top and bottom of $$\frac{(1+x)^5}{1+x^5}$$ by $1+x$ gives $$\frac{x^4+4x^3+6x^2+4x+1}{x^4-x^3+x^2-x+1}$$ Dividing top and bottom by $x^2$ gives $$\frac{x^2+\frac{1}{x^2}+4(x+\frac 1x)+6}{x^2+\frac{1}{x^2}-(x+\frac 1x)+1}$$ to have $$\frac{\left(x+\frac 1x\right)^2-2+4(x+\frac 1x)+6}{\left(x+\frac 1x\right)^2-2-(x+\frac 1x)+1}$$ Substituting $(1)$ into this gives $$\frac{(1+x)^5}{1+x^5}=\frac{\left(\frac{-35}{4}\right)^2-2+4(\frac{-35}{4})+6}{\left(\frac{-35}{4}\right)^2-2-(\frac{-35}{4})+1}=\frac{729}{1349}$$

Nice solution ! Thank you everyone. – user403160 Mar 10 '17 at 12:40 — user403160, Mar 10 '17 at 12:40

Mark · Answer 2 · 2017-03-09T16:35:23.173

EDIT: In short: yes, your answer is correct (you can ignore everything else here:)

$\frac{(1+x)^3}{1+x^3} = \frac{9}{13}$ implies $8 x = -35 \pm 3 \sqrt{129}$ from which you can compute $\frac{(1+x)^5}{1+x^5} = 729/1349$. So there is only one possible coprime pair $a,b$, making the minimality of $a+b$ an empty condition.

(Note: it may seem at first that you would get a cubic equation for $x$ but that's not true since the two $x^3$ terms cancel each other out. So all you have to do to find $x$ is to solve a quadratic equation. You did find this equation, at which point all that remained to be done is (a) solve it, and (b) substitute the result into $(1+x)^5/(1+x^5)$ which is not pleasant but not hard either).

PS. The quickest way to evaluate $(1+x)^5$ or $1+x^5$ at $x = $root of a quadratic equation $Q(x)$ is to divide those polynomials by $Q(x)$ with long division and to take the remainder.

Find the minimal $a+b$.

2 Answers2