I came across an interesting problem on one of the wiki pages on brilliant.org related to polynomials and the remainder theorem. The problem statement is as follows:
Find all polynomials $f(x)$ which satisfy the following conditions:
1. $f(x)$ is a monic polynomial
2. $f(x)$ has a degree 1000
3. $f(x)$ has integer coefficients
4. $f(x)$ divides $f(2x^3+x)$
If $f(x)$ has the property that $f(a*b)=f(a)*f(b)$ for all $a,\ b$, then condition 4 is satisfied as $2x^3+x$ can be factored as $x*(2x^2+1)$. So I tried to find all polynomials with the above property.
Then, any such polynomial must also satisfy $f(cx)=f(c)f(x)$ where $c$ is a constant..
Setting $c=0$, we get $f(0)=0$ and thus there is no constant term in $f(x)$, but then this leads us to just one possibility for $f(x)$ namely, $f(x)=x^{1000}$.
In this way, we get just one solution to the given problem.
But the solution to the problem is 501, meaning there are 501 such polynomials which satisfy the stated conditions. Where do the other solutions come from??
Link to the web page(site: brilliant.org) : https://brilliant.org/problems/it-isnt-easy-for-fx-to-divide-f2x3x/
Thanks for any answers!!
$f(x)$ can't be multiplicative, polynomials aren't multiplicative though the information given is not strong enough to find $f(x)$, but since $f(x)$ divides $f(2x^3+x)$, and we're determind to find $f(x)$
Then it's easy to assume that $f(x)$ has no constant term because $2x^3+x$ factors $x\cdot(2x^2+1)$, so that $f(x)$ can be divisible
this is a list of shapes $f(x)$ can take and there are $1000$ of them
$f(x) = x^{1000}$
$f(x) = x^{1000}+ax^{999}$
$f(x) = x^{1000}+ax^{999}+bx^{998}$
$f(x) = x^{1000}+ax^{999}+bx^{998}+cx^{997}$
$f(x) = x^{1000}+ax^{999}+bx^{998}+cx^{997}+dx^{996}$
$f(x) = x^{1000}+ax^{999}+bx^{998}+cx^{997}+dx^{996}+e^{995}$
$f(x) = x^{1000}+ax^{999}+bx^{998}+cx^{997}+dx^{996}+ex^{995}+...........+x$
Where $a,b,c,d,e \in Z$,