Find all polynomials $p(x)$ such that:
$$xp(x-1) = (x-30)p(x)$$
My solution:
We can see, if $x = 0$ then $p(0) = -0/29 = 0$ similarly, $p(29) = 0$ so our polynomial is $x(x-29)$.
I'd thought this was a simple question, but apparently the answer is:
$ax(x-1)(x-2)(x-3)\cdots(x-29)$, where a is any real number.
I have no idea how this is the answer
Your approach is almost correct. You are at half way.
You got $P(0)=P(29)=0$. Now consider, $P(1)$. From $$x\cdot P(x-1)=(x-30)P(x),$$ we have when $x=1$, $$1\cdot P(0)=(-29)P(1).$$ $$\implies P(1)=0.$$ Similarly, considering $x=2, 3, \ldots, 29$, you will get $P(2)=P(3)=\cdots=P(29)=0$. Hence your result follows.
Even your short answer is missing something -- you should multiply by an unknown constant because knowing the roots tells you nothing about the vertical scale of the polynomial.
The correct version of the argument you are attempting is this...
What happens when $x = 1$? Then $p(0) = -29 p(1)$. Since $p(0) = 0$, we have $p(1) = 0$.
We just got $p(1)$ on the right. To get $p(1)$ on the left, set $x = 2$. Then $2 p(1) = -28 p(2)$. Since $p(1) = 0$, we have $p(2) = 0$.
Repeating with $x = 3$, $4p(2) = 27 p(3)$ and so $p(3) = 0$.
...
Continuing, you eventually show all of $0$, $1$, ..., $29$ are roots. As in the first paragraph, you need a constant multiple since you have no way to get another value of the polynomial.
That seems to give you the answer you recite, but there is more to show. How do we know there aren't more roots? Suppose there were; for example, let $$ p(x) = a x(x-1)(x-2)\cdots (x-29) \cdot (x-100) \text{.} $$ Then the equation you start with forces $101$ is a root, which forces $102$ is a root, which forces ..., producing infinitely many roots. If you work through the details, you can show that the presence of any root other than those listed in the recited answer forces infinitely many more roots. Since no polynomial has infinitely many roots, there are no roots other than those in the recited answer.
So that only leaves repetitions of the $30$ roots we know about. Set $$ p(x) = a \prod_{k=0}^{29} (x-k)^{q_k} \text{.} $$
In $x p(x-1)$, the factors $(x-30)$ and $x$ appear with multiplicities $q_{29}$ and $1$, respectively. In $(x-30)p(x)$, with multiplicities $1$ and $q_0$, respectively. So $q_{29} = 1 = q_0$. Applying these henceforth, ...
In $x p(x-1)$, the factors $(x-29)$ and $x-1$ appear with multiplicities $q_{28}$ and $1$, respectively. In $(x-30)p(x)$, with multiplicities $1$ and $q_1$, respectively. So $q_{28} = 1 = q_1$. Applying these henceforth, ...
...
Continuing, we show all the $q_{k} = 1$, so all the roots have multiplicity one.
From $(x-30)p(x)=xp(x-1)$ and $p(0)=0$, we have $$(1-30)p(1)=1\cdotp(0)=0$$ i.e. $p(1)=0$. Also $$(2-30)p(2)=2p(1)=0$$ i.e. $p(2)=0$. Generally, if $p(k)=0$, for some positive integer $k$, then $$(k+1-30)p(k+1)=(k+1)p(k)=0$$ therefore, we have $$p(x)=x(x-1)(x-2)\cdots(x-29)g(x)$$ We claim $p(30)\ne 0$ else, if $p(30)=0$ then we will have $$p(31)=p(32)=\cdots=p(n)=\cdots=0$$ for all $n\geq30$ which implies $p\equiv0$.
You approach is correct. Indeed $p(0)=0$. Now substitute $x=1$. We get
$p(0) = -29p(1) = 0$ $\implies p(1) = 0$
Now substitute $x=2$ and in a similar fashion you obtain $p(2) = 0$. This continues until $x=30$ when $30p(29) = 0*p(30)$. Hence $x=0$ to $29$ are all roots which yields :
$p(x) = ax(x-1)(x-2)...(x-29)$
As you can judge from my comment on the solutions, they are all incomplete. What the analysis on $ p (0) = p(1) = \ldots p (29) = 0 $ indicates is only that
$$ p(x) = A(x) \times x (x-1)(x-2) \ldots (x-29), $$
where $A(x)$ is a polynomial.
When we substitute this back into the given equation, and divide by the common factors we get is
$$ A (x-1) = A(x). $$
This implies that $ A(x-1) = A(x) = A(x+1) = A(x+2) = \ldots$.
However, the only polynomial that takes on the same values at infinitely many points, is the constant polynomial. Thus $A(x) = a$ for some real number $a$.
This completes the solution.
