A nice problem with Polynomials
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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 $P(x)$.
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
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Re: A nice problem with Polynomials
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:
Solution:
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
Re: A nice problem with Polynomials
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,sourav das wrote: Solution:
Using this,deriving $k | a_m$ is an almost one-liner.Let $P(x)$ be a polynomial with integer coefficients and $a,b \in \mathbb{Z}$
Then \[a-b |P(a)-P(b)\]
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Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
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Re: A nice problem with Polynomials
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.
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.
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
Re: A nice problem with Polynomials
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.sourav das wrote: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 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.
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi