A nice problem with Polynomials

[sourav das]

A polynomial $P(x)$ of degree $n\geq 5$ with integer coefficients and $n$ distinct integer roots is given. Find all integer roots of $P(P(x))$ given that $0$ is a root of $P(x)$.

[sourav das]

I found this problem in an Algebra book and solved it using Number theory. And the given solution was also quite long and used many tricks. But i solved it quite in a simple way. So I'm little bit confused. I'm giving my solution. Please help me by informing me if you find any bug.

Solution:

As $0$ is a root of$ P(x)$, let $P(x)=cx\prod_{i=2}^{n} (x-a_i)$ where $c$ is a constant and $a_i$ are non $0$ integer roots of $P(x)$. Now as $0$ is a root of $P(x)$ , all integer roots of $P(x)$ are also roots of $P(P(x))$. We claim no other integer root exists.

On the contrary, let $k$ not equal to $0$ is a root of $P(P(x))$. Then $P(k)\in \{a_i: 1<i \leq n \}$ as $P(k)$ must be a root of $P(x)$. Now let $P(k)=a_m$. So $ck(k-a_m)\prod (k-a_i)=a_m$....(i). So $k|a_m$. Let $a_m=kb_m$. Plugging this in (i) we'll get $ck(1-b_m)\prod(k-a_i)=b_m$. But g.c.d.$(1-b_m,b_m)=1$. So $b_m=o$ or $2$. But $b_m$ cannot be $0$ as then $a_m=0$ contradiction. Letting $b_m=2$ it'll lead us to, $c(k-a_m)\prod(k-a_i)=2$. But as $n\geq 5$, and all roots of $P(x)$ are distinct; there are at least distinct $(k-a_p),(k-a_q),(k-a_r),(k-a_m)$ 4 factors in L.H.S . So these will give at least 1 more factor greater that 1 including $(2,1,-1)$ or $(-2,1,-1)$. A contradiction as R.H.S. is only $2$.

So all integer roots of $P(x)$ are only integer roots of $P(P(x))$

[*Mahi*]

Solution:

As $0$ is a root of$ P(x)$, let $P(x)=cx\prod_{i=2}^{n} (x-a_i)$ where $c$ is a constant and $a_i$ are non $0$ integer roots of $P(x)$. Now as $0$ is a root of $P(x)$ , all integer roots of $P(x)$ are also roots of $P(P(x))$. We claim no other integer root exists.

On the contrary, let $k$ not equal to $0$ is a root of $P(P(x))$. Then $P(k)\in \{a_i: 1<i \leq n \}$ as $P(k)$ must be a root of $P(x)$. Now let $P(k)=a_m$. So $ck(k-a_m)\prod (k-a_i)=a_m$....(i). So $k|a_m$. Let $a_m=kb_m$. Plugging this in (i) we'll get $ck(1-b_m)\prod(k-a_i)=b_m$. But g.c.d.$(1-b_m,b_m)=1$. So $b_m=o$ or $2$. But $b_m$ cannot be $0$ as then $a_m=0$ contradiction. Letting $b_m=2$ it'll lead us to, $c(k-a_m)\prod(k-a_i)=2$. But as $n\geq 5$, and all roots of $P(x)$ are distinct; there are at least distinct $(k-a_p),(k-a_q),(k-a_r),(k-a_m)$ 4 factors in L.H.S . So these will give at least 1 more factor greater that 1 including $(2,1,-1)$ or $(-2,1,-1)$. A contradiction as R.H.S. is only $2$.

So all integer roots of $P(x)$ are only integer roots of $P(P(x))$

My solution to this is quite similar and as far as I know ,it is alright. But there is one thing, I used this uncommon (and may be self made, I don't know) lemma,

Let $P(x)$ be a polynomial with integer coefficients and $a,b \in \mathbb{Z}$
Then \[a-b |P(a)-P(b)\]

Using this,deriving $k | a_m$ is an almost one-liner.

[sourav das]

Actually i need the part g.c.d.$(b_m,1-b_m)=1$ That's why i just do it in equations.

By the way, I think it's a great lemma. If i'm not wrong the actual formation is:

If $P(x)$ is a polynomial with integer co-efficients.If $a\equiv b$ (mod $n$) then $P(a)\equiv P(b)$ (mod $n$)

And i think you set $a=k$ and $b=0$. Really a nice one.

[*Mahi*]

By the way, I think it's a great lemma. If i'm not wrong the actual formation is:

If $P(x)$ is a polynomial with integer co-efficients.If $a\equiv b$ (mod $n$) then $P(a)\equiv P(b)$ (mod $n$)

This form of the lemma is nowhere as useful as it is when it is in the other (divisibility) form, that's why I called it different.
And also, when you've proved $k|a_m$, use this lemma to prove $k-a_m|a_m$, and then $k=0,2$ is easy to derive.