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I want to know how to answer this question:

Find all polynomial in $mathbbR[X]$ such that the derivative $P' | P$.

My effort:

We know that

• An $n$ degree polynomial has $n$ roots in $mathbbC$.




• $P(x) = P'(x) cdot (ax+b).$

Thus we can suppose that
$$P' = b(x-a_1)(x-a_2)dots(x-a_n-1),$$
and
$$P = c(x-a_1)(x-a_2)dots(x-a_n-1) (x-a_n)$$
where $b, cin mathbbR$ and $a_1,dots,a_n in mathbbC$.
Take the derivative we have
$$P' = c(x-a_2)(x-a_3)dots(x-a_n)+c(x-a_1)(x-a_2)dots(x-a_n)+ dots + c(x-a_2)(x-a_3)dots(x-a_n-1).$$
Since $P'(a_1) = 0$, we then have $c(a_1-a_2)dots(a_1-a_n) = 0$, thus $xin a_2,dots, a_n$. Suppose that $a_1=a_2$.
Then $$P = c(x-a_1)^2(x-a_3)dots(x-a_n-1) (x-a_n).$$
Hopefully at some point, I can get $P = c(x-a_1)^n$.

Write $P(x) = P'(x) L(x)$, where $L$ has degree $1$.

Write $P(x) = L(x)^m Q(x)$, where $Q$ is coprime with $L$.

Then $P' = m L^m-1 L' Q + L^m Q'$ and so $L^m Q = P = P'L = m L^m L' Q + L^m+1 Q'$.
Therefore, $Q$ divides $ L^m+1 Q'$. Since, $Q$ is coprime with $L$, $Q$ must divide $Q'$, and this happens iff $Q$ is constant.

Thus, $P=q L^m$.

Let $$P(x) = a_nx^n+....+a_1x+a_0,$$ then we have $$ a_nx^n + ...+a_1x+a_0 = (a_ncdot ncdot a) x^n+...+(aa_1+2a_2b)x+a_1b$$ Since $a_nneq 0$ we get $an=1$, so $a=1/n$

Write $$ P'(x) over P(x) = 1over ax+b$$

Then $$a(ln P(x))' = (ln(ax+b))'$$ and thus $$aln(P(x) = ln(ax+b) + c$$

So $$P(x) = (ax+b)^1over a e^c = (ax+b)^ne^c$$

For a polynomial $P$ of degree $n$, we always have

$$P'(x)over P(x)=1over x-r_1+1over x-r_2+cdots+1over x-r_n$$

where $r_1,r_2,ldots,r_n$ are the roots of $P$. If, at the same time, we have $P'mid P$, then $P(x)=1over n(x-r)P'(x)$ for some $r$ (since $P'$ is of degree $n-1$ with lead coefficient that is $n$ times the lead coefficient of $P$). This gives us

$$nover x-r=1over x-r_1+1over x-r_2+cdots+1over x-r_n$$

If we do not have $r_1=r_2=cdots=r_n=r$, then the right hand side has a singularity where the left hand side does not. Thus if $P'mid P$, we must have $r_1=r_2=cdots=r_n=r$, and so $P$ must have the form $P(x)=a(x-r)^n$. We also need $a,rinmathbbR$ in order for $P$ to have real coefficients.

Remark: The key equality can be proved by induction: If $P(x)=(x-r_n)Q(x)$ with

$$Q'(x)over Q(x))=1over x-r_1+1over x-r_2+cdots+1over x-r_n-1$$

then

$$beginalign
P'(x)over P(x)
&=((x-r_n)Q(x))'over(x-r_n)Q(x)\
&=(x-r_n)Q'(x)+Q(x)over(x-r_n)Q(x)\
&=Q'(x)over Q(x)+1over x-r_n\
&=1over x-r_1+1over x-r_2+cdots+1over x-r_n-1+1over x-r_n
endalign$$

(with the base case, $P(x)=a(x-r_1)$, being straightforward to check).

Assume $P$ is of degree $n$. Then $P'$ is of degree $n-1$. Then of course, $P'| P$, means that there is a polinomial $Q$ of degree 1 such that $P'Q = P$. (Otherwise the degree of $QP'$ would not be the same as the one of $P$)

Let's write $P = a_n x^n + ... + a_1x + a_0$. Then $P' = n a_n x^n-1 + (n-1)a_n-1x^n-2+... + a_1$

We can write $Q$ as $bx + c$, so $Q P' = (bx+c)(n a_n x^n-1 + (n-1)a_n-1x^n-2+... + a_1) = $ $= a_n n b x^n + (a_n-1(n-1)b + na_n c)x^n-1+... + (a_i-1 (i-1)b+a_i ic)x^i-1 + (a_1 b + a_2 2 c )x+a_1 c $

Then, since $QP' = P$, we must have
begincases a_n = a_n n b \ ... \ a_i-1 = a_i-1 (i-1) b + a_i i c \ ... \ a_0 = a_1c endcases

For the first equation we know we must have $b = frac1n$ and from the last $c = fraca_0a_1$

Then, if all the other $a_i-1 = a_i-1 (i-1) b + a_i i c$ for $i = 1... n-1$ are good with that, we reach our goal. That means they have to solve $a_i-1 = a_i-1 (i-1) frac1n + a_i i fraca_0a_1$

That is $a_i-1 (1 - fraci-1n) = a_i i fraca_0a_1$, which is $a_i-1 = a_i i fraca_0a_1fracnn-i+1= a_i fraca_0a_1fracinn-i+1$

That gives us a way to find, with free $a_0, a_1, a_n$ all the others $a_i$. However, if we apply that to $a_1$, it gives us $a_1=a_2 fraca_0a_1fracnn $, which means $a_1 ^2 = a_2 a_0$