If you don't have time or don't feel like readings lot (or what ever reason you may have) there is an abridged version at the bottom that is much easier to read.

In this note, I will prove that for all polynomials $f(x)=a_n x^n+a_{n-1}x^{n-1}+...+a_0$, with $a_n ~ and ~ a_0 \neq 0$ and x=0 is not a root, that after being divided by GCD of the coefficients ($g$) or multiplied by a factor of $k$ such that the new coefficients are both integral and co-prime, the function with the smallest possible integral $inverse$ roots has the form $f'(x)=p(a_0x^n+a_1x^{n-1}...+a_n)$ where p is a constant and $p=1$ if and only if $f(x)$ has all group wise co-prime integer coefficients.

(In the short, I'm proving that the coefficients are reversed if the roots are inverted (AKA: $r_1$ becomes $\dfrac{1}{r_1}$)

By Vieta's, we have $-\dfrac{a_{n-1}}{a_n}=r_1+r_2+r_3 \dots+ r_n$ for $f(x)$

Now, assume that we didn't know that $f'(x)$ had the reverse coefficients of $f(x)$

Thus $f'(x)=b_nx^n+b_{n-1}x^{n-1}...+b_0$

However, we know that the roots of $f'(x)$ are $\dfrac{1}{r_i}$ where $0\leq i \leq n$.

For $f'(x)$, we have $-\dfrac{b_{n-1}}{b_n}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3} \dots+ \dfrac{1}{r_n}$

If we multiply both the top and bottom of every fraction on the R.H.S. by every root other than the one in its denominator, we will have $-\dfrac{\left(\dfrac{a_1}{a_n}\right)}{\left(\dfrac{a_0}{a_n}\right)}\Rightarrow -\dfrac{a_1}{a_0}$ (This is too ugly to show in "proper" math format here. So I'll show a simpler and proper version below)

$\therefore -\dfrac{a_1}{a_0}=-\dfrac{b_{n-1}}{b_n}$

It can be easily observed that $b_n=p(a_0)$ where p is a constant and $b_{n-1}=p(a_1)$.

This process can be repeated to show that $\dfrac{a_{i+1}}{a_{i}}=\dfrac{b_{n-(i+1)}}{b_{n-(i)}}$ where $0\leq i < n$ (NOTE: this is the one time where $i<n$ rather than $i \leq n$ because there is no $b_{-1}$ nor $a_{n+1}$

We can see that the ratio's of the $a_i$ is equal to the ratio's of the $b_i$ in reverse order. Going back to our other equation $\dfrac{a_1}{a_0}=\dfrac{b_{n-1}}{b_n}$. Assuming that $f(x)$ has all $group ~ wise$ co-prime coefficients, $\dfrac{a_1}{a_0}$ could be reducible by a factor of h. However, this leads to two contradictions: 1. every other ratio would have to be reduced by h or $f(x)$ would be changed. 2. If every other coefficient was reduced then some would become non-integers if they were all group wise co-prime (if they weren't then this would contradict a condition earlier).

Also since we're looking for an $f'(x)$ with the smallest possible co-prime integer coefficients, $\dfrac{b_{n-1}}{b_n}$ should be irreducible. These statements imply that $b_{n-1}=a_1$ and $b_n=a_0$.

Once again, we can use this logic to prove that $a_i=b_{n-i}$ where $0\leq i \leq n$.

Thus we have proved that for integral coefficients of $f(x)$, $f'(x)=a_0x^n+a_1x^{n-1}...+a_n$

If $f(x)$ doesn't have integer coefficients, then we can multiply by a factor of $p$ such that the coefficients become integral, thus our final result is $f'(x)=p(a_0x^n+a_1x^{n-1}...+a_n)$.

$-\dfrac{b_{n-1}}{b_n}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3} \dots+ \dfrac{1}{r_n}=\displaystyle \sum_{i=1}^{n} \left(\dfrac{\left( \displaystyle \sum_{i=1}^{n} (r_i) \right)}{r_i}\right)\cdot \dfrac{1}{\left( \displaystyle \sum_{i=1}^{n} (r_i) \right)}$

$\Rightarrow \displaystyle \sum_{i=1}^{n} \left(\dfrac{\left(\dfrac{a_0}{a_n}\right)}{r_i}\right)\cdot\dfrac{1}{\left(\dfrac{a_0}{a_n}\right)}= \displaystyle \sum_{i=1}^{n} \left(\dfrac{1}{a_n}\cdot \dfrac{a_0}{r_i}\right)\cdot\dfrac{a_n}{a_0}$

$\Rightarrow \left(\dfrac{1}{a_n}\cdot (a_1)\right)\cdot\dfrac{a_n}{a_0}=\dfrac{a_1}{a_0}$

Polynomial $ax^3+bx^2+cx+d$ has roots p,q,r.

$-\dfrac{b}{a}=p+q+r$

$\dfrac{c}{a}=pq+qr+rp$

$-\dfrac{d}{a}=pqr$

Polynomial $hx^3+gx^2+jx+k$ has roots $\dfrac{1}{p}, \dfrac{1}{q}, \dfrac{1}{r}$

-\dfrac{g}{h}=\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}=\dfrac{(pq)}{(pq)r}+\dfrac{(rp)}{(rp)q}+\dfrac{(pq)}{(pq)r}=\dfrac{pq+qr+rp}{pqr}=\Rightarrow -\dfrac{\left(\dfrac{c}{a}\right)}{\left(\dfrac{d}{a}\right)}=\boxed{\color \pink{-\dfrac{c}{d}}}

\dfrac{j}{h}=\dfrac{1}{pq}+\dfrac{1}{qr}+\dfrac{1}{rp}=\dfrac{(r)}{(r)pq}+\dfrac{p}{(p)qr}+\dfrac{(q)}{(q)rp}=\dfrac{p+q+r}{pqr}=\dfrac{\left(\dfrac{b}{a}\right)}{\left(\dfrac{d}{a}\right)}=\boxed{\color \green{\dfrac{b}{d}}}

-\dfrac{k}{h}=\dfrac{1}{pqr}=\dfrac{1}{\left(-\dfrac{d}{a}\right)}=\boxed{\color \red{-\dfrac{a}{d}}}

Assuming a monic polynomial (we can divide by h because dividing by a constant won't change the roots)

$hx^3+gx^2+jx+k$ has equivalent roots to x^3+\dfrac{g}{h}x^2+\dfrac{j}{h}x+\dfrac{k}{h}=x^3+\color \pink{\dfrac{c}{d}}x^2+\color \green{\dfrac{b}{d}}x+\color \red{\dfrac{a}{d}}

Multiplying through to rationalize the denominators by d.

$\boxed{dx^3+cx^2+bx+a}$

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## Comments

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TopNewestWhy not tranform the equation by the substitution $x=\dfrac{1}{y}$ Its a 1 line proof after that.

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That transformation is certainly a much quicker approach. Can you add more details about what you did?

For a problem along the same lines, check out Squaring the roots of a polynomial.

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I'm not sure exactly what you mean, but by your substitution, since we're solving for roots, isn't $x=\infty$

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Why $x=\infty$?

Aren't we just inverting the roots?

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$\x=\dfrac{1}{y}$. And since we're finding roots y=0

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I just used $y$ to take a different variable. I did not mean the $y$ in $y=f(x)$

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Note: You should state that $x = 0$ is not a solution, or that equivalently $a_0 \neq 0$. This would also justify allowing you to divide by $a_0$.

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Ahh, thanks, thats a good point.

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Thanks bruh

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Chee Bra.

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