Prove that if $Q(x) \mid P(x)$ and $P(x) \mid Q(x)$ then $Q(x)=cP(x)$ where $c≠0$ is a constant

Question

I am having hard time proving this theorem in polynomial algebra. I would appreciate any tips or advice on approaching this task.

Many thanks in advance.

Answer 1

Hints:

1) If $Q$ divides $P$, then by definition there exists a polynomial $R$ such that $P=R\cdot Q$.

2) $deg(R\cdot Q)=deg(R)+deg(Q)$, where $deg$ is the degree of the polynomial.

Answer 2

Hint.

$$ P(x) = u(x)Q(x)\\ Q(x) = v(x)P(x) $$

then

$$ P(x) = u(x)v(x)P(x)\Rightarrow \mbox{deg}(u(x)v(x))=0\Rightarrow\mbox{deg}(u(x)) = \mbox{deg}(v(x))= 0 $$

Answer 3

By definition of polynomial division only a polynomial of lower or equal degree can divide a polynomial of higher or equal degree. Here however both polynomials divide each other hence their degree must be the same. As for the coefficients ... if the corresponding coefficients are not the same or multiple of each other (for these two same degree polynomials) then any of the two divisions will give a remainder that would be function of x leading to reduction or gain of degree for one of the polynomials in the division process thus contradicting the correct 'same degree' hypothesis...hope this helps.