Let $f(x)$ be a polynomial with integer coefficients. Assume that $3$ divides the value $f(n)$ for each integer $n.$ Prove that when $f(x)$ is divided by $x^3-x, $ the remainder is of the form $3r(x),$ where $r(x)$ is a polynomial with integer coefficients.
An Edited Solution is presented below following the suggestion of @dxiv
My solution goes like this:
If $f(x)$ is divided by $x^3-x, $ then, by division algorithm we have $f(x)=(x^3-x)g(x)+p(x),$ where $g(x),p(x)$ are the quotient and remainder respectively when $f(x)$ is divided by $x^3-x .$ We note that as $f(x)$ and $x^3-x, $ is a polynomial with integer coefficients, so $p(x),g(x)$ should also have integer coefficients. Now, $3|f(x)$ for $x\in \Bbb N$ and $$3!|(x-1)x(x+1)\implies 3|x(x-1)(x+1)$$ (since product of three consecutive numbers is divisible by $3! $). Hence, $$3| f(x)-x(x+1)(x-1)g((x)\implies 3|p(x),$$( when $x\in\Bbb N$) and this means $p(x)=3r(x)$ , where $r(x)$ is a polynomial with integer coefficients.
However, I concluded "$3|p(x),$ when $x\in \Bbb N$." Is this a justified (proved) claim? I feel something might be wrong in this proof. But I don't understand it. I think, the real picture, that gives me such an uncanny feeling is that the underlying assertion/fact I used is that : If $f(x)$ is a polynomial with integer coefficients such that $p|f(x)$ for all natural number $x$ and $p$ is a prime, then $p(x)$ is of the form $3q(x)$, where $q(x)$ also, is a polynomial with integer coefficients. I probably used the above fact, unknowingly but if this assertion that I utilised in the above solution to conclude, if at all is a valid one, I could figure this holding true only intuitively. I want to know, if this is really the case ( i.e the above assertion is used unknowingly to solve the problem or not) ? All in all, does my solution looks justified ?
Following @dxiv 's suggestion, I proceeded like this:
If $f(x)$ is divided by $x^3-x, $ then, by division algorithm we have $f(x)=(x^3-x)g(x)+p(x),$ where $g(x),p(x)$ are the quotient and remainder respectively when $f(x)$ is divided by $x^3-x .$ We note that as $f(x)$ and $x^3-x, $ is a polynomial with integer coefficients, so $p(x),g(x)$ should also have integer coefficients. Now, $3|f(x)$ for $x\in \Bbb N$ and $$3!|(x-1)x(x+1)\implies 3|x(x-1)(x+1)$$ (since product of three consecutive numbers is divisible by $3! $). Hence, $$3| f(x)-x(x+1)(x-1)g((x)\implies 3|p(x),$$( when $x\in\Bbb N$) coefficients. We now observe, as $x^3-x$ has degree $3$ so, $p(x)$ has a degree of atmost $2.$ This implies, from @Soumik Mukherjee's suggestion, that $3|f(1),f(2),f(3).$ We asumme $p(x)=ax^2+bx+x$, where $a$ might/might not be equal to $0$, depending on what the degree of $p(x)$ is. Now, we have, $3|p(1),p(2),p(3).$ So, we can say $3|a+b+c,4a+2b+c,9a+3b+c$ due to which, we have $3|a,b,c.$ Assuming $a=3q,b=3w,c=3e,$ we have $p(x)=3qx^2+3wx+3e=3(qx^2+wx+e)=3r(x)$, where $r(x)=qx^2+wx+e$ and $w,x,e\in \Bbb Z,$ for which we conclude $r(x)$ is a polynomial with integer coefficient. This completes the proof.
Does this modified solution valid in all it's argument ?