roots - A simple zeroes polynomial which constructs another one

Let $P(X) = (X-x_1)\ldots(X-x_N)$ be a complex polynomial with simple roots.

I define
$$Q(X) = P(X-a)-bP(X),$$

with $a\in\mathbb{C}$ and $b\neq 1$ so that $Q(X)$ is also a polynomial of degree $N$.

Let me note $y_1,\ldots,y_N$ the roots of $Q(X)$. I would like to prove that

• $Q$ has simple roots : $\forall i,j, \quad y_i \neq y_j$


• one cannot shifts from a root to another by the use of $a$ : $\forall i,j,\quad y_i \neq y_j \pm a$

I am not sure this is the case for all $a$ and $b$, but in the context of my problem it seems $Q$ has to verify these conditions.

I thought it would be a simple exercise, but I keep struggling on it.

Any help is much appreciated ! :)

Answer

It's not true.

As a counterexample, letting

• $P(x)=(x+2)(x-2)$$\\[4pt]$
  
• $a=5$$\\[4pt]$
  
• $b=-4$
  

we get

$\;\;\;Q(x)=5(x+1)^2$.

Staying with the case $n=2$, let $P(x)=(x-r)(x-s)$ with $r\ne s$.

Then for $b\ne 1$, the polynomial
$$Q(x)=P(x-a)-bP(x)$$
has simple roots if and only if either$\;b=0\;$or
$$a^2\ne -\frac{(b-1)^2(r-s)^2}{4b}$$

Let's try an example with $n=3$ . . .

Let $P(x)=(x-1)x(x+1)$.

Then for $b\ne 1$, the polynomial
$$Q(x)=P(x-a)-bP(x)$$
has simple roots if and only if
$$4b^4+(36a^2-16)b^3+(-27a^6+108a^4-72a^2+24)b^2+(36a^2-16)b+4$$
is nonzero.

The results of that example suggest that for the general case, trying to find usable necessary and sufficient conditions on $a,b$ for $Q(x)$ to have simple roots is not likely to succeed.

As a more reasonable goal, one might try to find sufficient conditions for $Q(x)$ to have simple roots, expressed in terms of inequalities relating $|a|,|b|$.

For example, if $a$ is fixed, then $Q(x)$ will have simple roots if

• $b\;$is sufficiently close to $0$.
  

  
  or
  

• $|b|\;$is sufficiently large.